From the points A And B, the distance between which is l, at the same time two bodies began to move towards each other: the first with a speed v 1, second - v 2. Determine how long after they will meet and the distance from the point A to their meeting place. Solve the problem also graphically.
Solution
1st method:
Dependence of body coordinates on time:
At the moment of meeting, the coordinates of the bodies will coincide, i.e. This means that the meeting will occur after a time from the beginning of the movement of the bodies. Find the distance from the point A to the meeting place as .
2nd method:
The velocities of the bodies are equal to the tangent of the angle of inclination of the corresponding graph of the coordinates versus time, i.e., , . The moment of the meeting corresponds to a dot C graph intersections.
After what time and where would the bodies meet (see problem 1) if they were moving in the same direction A→B, and from the point B the body began to move through t 0 seconds after it starts moving from the point A?
Solution
Graphs of the dependence of body coordinates on time are shown in the figure.
Based on the figure, let’s create a system of equations:
Having solved the system for t C we get:
Then the distance from the point A to the meeting point:
.
A motorboat travels the distance between two points A And B along the river in time t 1 = 3 hours, and the raft - in time t= 12 hours. What time t 2 will spend powerboat on the way back?
Solution
Let s— distance between points A And B, v is the speed of the boat relative to the water, and u— flow speed. Expressing distance s three times - for a raft, for a boat moving with the current, and for a boat moving against the current, we obtain a system of equations:
Having solved the system, we get:
A metro escalator takes a person walking down it in 1 minute. If a person walks twice as fast, he will descend in 45 seconds. How long does it take for a person standing on an escalator to descend?
Solution
Let us denote by the letter l escalator length; t 1 — descent time of a person walking at speed v; t 2 — descent time of a person walking at speed 2 v; t— time of descent of a person standing on the escalator. Then, having calculated the length of the escalator for three different cases (a person walks at a speed v, at speed 2 v and stands motionless on the escalator), we obtain a system of equations:
Having solved this system of equations, we get:
A man runs along an escalator. The first time he counted n 1 = 50 steps, the second time, moving in the same direction at three times the speed, he counted n 2 = 75 steps. How many steps would he count on a stationary escalator?
Solution
Since with increasing speed the person counted large quantity step, which means the directions of the speeds of the escalator and the person coincide. Let v— speed of a person relative to the escalator, u— escalator speed, l- length of the escalator, n— the number of steps on a stationary escalator. The number of steps that fit in a unit length of the escalator is equal to n/l. Then the time a person spends on the escalator when he moves relative to the escalator at a speed v equals l/(v+u), and the distance traveled along the escalator is equal to vl/(v+u). Then the number of steps counted on this path is equal to . Similarly, for the case when the speed of a person relative to the escalator is 3 v, we will receive .
Thus, we can create a system of equations:
Eliminating the attitude u/v, we get:
Between two points located on a river at a distance s= 100 km from one another, there is a boat cruising, which, going with the flow, covers this distance in time t 1 = 4 hours, and against the current - for the time t 2 = 10 hours. Determine the speed of the river. u and speed of the boat v regarding water.
Solution
Expressing distance s twice, for a boat going with the current and for a boat going against the current, we get a system of equations:
Solving this system, we get v= 17.5 km/h, u= 7.5 km/h.
A raft passes by the pier. At this moment in a village located at a distance s 1 = 15 km from the pier, a motor boat departs down the river. She reached the village in time t= 3/4 hours and, turning back, met the raft at a distance s 2 = 9 km from the village. What is the speed of the river current and the speed of the boat relative to the water?
Solution
Let v- speed of the motor boat, u— river flow speed. Since from the moment the motor boat departs from the pier until the moment the motor boat meets the raft, the same time will obviously pass for both the raft and the motor boat, the following equation can be drawn up:
where on the left is the expression of the time elapsed until the moment of meeting for the raft, and on the right - for the motor boat. Let's write the equation for the time it took the motor boat to cover the distance s 1 from the pier to the village: t=s 1 /(v+u). Thus, we obtain a system of equations:
Where do we get it from? v= 16 km/h, u= 4 km/h.
A column of troops during a march moves at speed v 1 = 5 km/h, stretching along the road for a distance l= 400 m. The commander, located at the tail of the column, sends a cyclist with an order to the lead detachment. The cyclist sets off and rides at speed v 2 = 25 km/h and, having completed the assignment on the move, immediately returns back at the same speed. After how long t did he return after receiving the order?
Solution
In the reference frame associated with the column, the speed of the cyclist when moving towards the lead column is equal to v 2 -v 1, and when moving back v 2 +v 1 . That's why:
Simplifying and substituting numeric values, we get:
.
Car width d= 2.4 m, moving at speed v= 15 m/s, was pierced by a bullet flying perpendicular to the movement of the car. The displacement of the holes in the walls of the car relative to each other is equal to l= 6 cm. What is the speed of the bullet?
Solution
Let us denote by the letter u bullet speed. The flight time of a bullet from wall to wall of the car is equal to the time it takes the car to travel the distance l. Thus, we can create an equation:
From here we find u:
.
What is the speed of the drops v 2 vertically falling rain, if the driver of a car notices that raindrops do not leave a mark on the rear window, tilted forward at an angle α = 60° to the horizon when the vehicle speed v 1 more than 30 km/h?
Solution
As can be seen from the figure,
To ensure that raindrops do not leave a mark on the rear window, it is necessary that the time it takes for the drop to travel the distance h was equal to the time during which the car will go the distance l:
Or, expressed from here v 2:
On the street it's raining. When will a bucket in the back of a truck be filled? faster with water: when the car is moving or when it is stationary?
Answer
Same.
At what speed v and at what course should the plane fly so that in time t= 2 hours fly exactly to the North way s= 300 km if during the flight the wind blows from the north-west at an angle α = 30° to the meridian with speed u= 27 km/h?
Solution
Let's write the system of equations according to the figure.
Since the plane must fly due north, the projection of its speed onto the axis Oy v y is equal y- wind speed component u y.
Having solved this system, we find that the plane should head northwest at an angle of 4°27" to the meridian, and its speed should be 174 km/h.
Moves along a smooth horizontal table at speed v Black board. What shape of mark will be left on this board by chalk thrown horizontally at speed u perpendicular to the direction of movement of the board, if: a) the friction between the chalk and the board is negligible; b) is friction high?
Solution
The chalk will leave a mark on the board, which is a straight line making an angle arctan( u/v) with the direction of movement of the board, i.e. it coincides with the direction of the sum of the speed vectors of the board and chalk. This is true for both case a) and case b), since the friction force does not affect the direction of movement of the chalk, since it lies on the same straight line with the velocity vector, it only reduces the speed of the chalk, so the trajectory in case b) may not reach the edge of the board.
The ship leaves the point A and goes at speed v, making an angle α with line AB.
At what angle β to the line AB should have been released from the point B a torpedo to hit a ship? The torpedo must be released at the moment when the ship was at the point A. The speed of the torpedo is u.
Solution
Dot C in the picture this is the meeting point between the ship and the torpedo.
A.C. = vt, B.C. = ut, Where t— time from the start to the moment of the meeting. According to the sine theorem
From here we find β :
.
To the slider, which can move along the guide rail,
attached cord threaded through the ring. The cord is selected at a speed v. At what speed u the slider moves at the moment when the cord makes an angle with the guide α ?
Answer and solution
u = v/cos α.
In a very short period of time Δt the slider moves a distance AB = Δl.
During the same period of time, the cord is selected to a length A.C. = Δl cos α (angle ∠ ACB can be considered right, since the angle Δα very small). Therefore we can write: Δl/u = Δl cos α /v, where u = v/cos α , which means that the speed of rope retraction is equal to the projection of the speed of the slider onto the direction of the rope.
Workers lifting loads
pull ropes at the same speed v. What speed u has a load at the moment when the angle between the ropes to which it is attached is equal to 2 α ?
Answer and solution
u = v/cos α.
Projection of load speed u the direction of the rope is equal to the speed of the rope v(see problem 15), i.e.
u cos α = v,
u = v/cos α.
Rod length l= 1 m articulated with couplings A And B, which move along two mutually perpendicular slats.
coupling A moves with constant speed v A = 30 cm/s. Find speed v B couplings B at the moment when the angle OAB= 60°. Taking the moment when the clutch A was at the point O, determine the distance O.B. and clutch speed B as a function of time.
Answer and solution
v B= v Actg α
= 17.3 cm/s; , 
.
At any point in time, velocity projections v A and v B ends of the rod
on the axis of the rod are equal to each other, since otherwise the rod would have to be shortened or lengthened. So we can write: v A cos α = vB sin α . Where vB = v A ctg α .
At any time for a triangle OAB The Pythagorean theorem is true: l 2 = O.A. 2 (t) + O.B. 2 (t). Let's find it from here O.B.(t): . Because the O.A.(t) = v A t, then we will finally write down the expression for O.B.(t) So: .
Since ctg α
at any moment is equal to O.A.(t)/OB(t), then we can write an expression for the dependence vB from time: .
The tank moves at a speed of 72 km/h. At what speed do they move relative to the Earth: a) top part caterpillars; b) Bottom part caterpillars; c) the point of the caterpillar, which is at this moment moves vertically relative to the tank?
Answer and solution
a) 40 m/s; b) 0 m/s; c) ≈28.2 m/s.
Let v- speed is the speed of the tank relative to the Earth. Then the speed of any point on the track relative to the tank is also equal to v. The speed of any point on the track relative to the Earth is the sum of the vectors of the speed of the tank relative to the Earth and the speed of the point on the track relative to the tank. Then for case a) the speed will be equal to 2 v, for b) 0, and for c) v.
1. The car drove the first half of the journey at a speed v 1 = 40 km/h, second - at speed v 2 = 60 km/h. Find average speed along the entire path traveled.
2. The car drove half the distance at speed v 1 = 60 km/h, the rest of the way he walked at speed half the time v 2 = 15 km/h, and the last section at speed v 3 = 45 km/h. Find the average speed of the car along the entire journey.
Answer and solution
1. v av =48 km/h; 2. v av = 40 km/h.
1. Let s- all the way, t- time spent to cover the entire path. Then the average speed along the entire path is s/t. Time t consists of the sum of the time intervals spent covering the 1st and 2nd halves of the journey:
.
Substituting this time into the expression for the average speed, we get:
.(1)
2. The solution to this problem can be reduced to solution (1.), if you first determine the average speed in the second half of the path. Let's denote this speed vср2 , then we can write:
Where t 2 - time spent to overcome the 2nd half of the journey. The path traveled during this time consists of the path traveled at speed v 2, and the distance covered at speed v 3:
Substituting this into the expression for vср2 , we get:
.
.
For the first half of the journey, the train traveled at a speed of n=1.5 times greater than the second half of the path. Average speed of the train along the entire journey v cp = 43.2 km/h. What are the speeds of the train at first ( v 1) and second ( v 2) half way?
Answer and solution
v 1 =54 km/h, v 2 =36 km/h.
Let t 1 and t 2 - time for the train to travel through the first and second half of the journey, respectively, s- the entire distance covered by the train.
Let's create a system of equations - the first equation is an expression for the first half of the path, the second - for the second half of the path, and the third - for the entire path traveled by the train:
By making a substitution v 1 =nv 2 and solving the resulting system of equations, we obtain v 2 .
Two balls began to move simultaneously and at the same speed along surfaces having the shape shown in the figure.
How will the speeds and times of movement of the balls differ by the time they arrive at the point? B? Ignore friction.
Answer and solution
The speeds will be the same. The first ball will take longer to move.
The figure shows approximate graphs of the movement of the balls.
Because the paths traveled by the balls are equal, then the areas of the shaded figures are also equal (the area of the shaded figure is numerically equal to the distance traveled), therefore, as can be seen from the figure, t 1 >t 2 .
The plane flies from the point A to point B and returns back to point A. The speed of an airplane in calm weather is v. Find the ratio of the average speeds of the entire flight for two cases when the wind blows during the flight: a) along the line AB; b) perpendicular to the line AB. The wind speed is u.
Answer and solution
Airplane flight time from point A to point B and back when the wind blows along the line AB:
.
Then the average speed in this case is:
.
If the wind blows perpendicular to the line AB, the aircraft's velocity vector must be directed at an angle to the line AB so as to compensate for the influence of wind:
The round-trip flight time in this case will be:
Aircraft flight speed to point B and vice versa are the same and equal:
.
Now we can find the ratio of the average speeds obtained for the cases considered:
.
Distance between two stations s= 3 km a metro train travels at average speed v avg = 54 km/h. At the same time, it takes time to accelerate t 1 = 20 s, then goes evenly for some time t 2 and it takes time to slow down to a complete stop t 3 = 10 s. Graph the speed of the train and determine the highest speed of the train v Max.
Answer and solution
The figure shows a graph of the speed of a train.
The distance traveled by the train numerically equal to area figures, limited by schedule and the time axis t, so we can write the system of equations:
From the first equation we express t 2:
,
then from the second equation of the system we find v Max:
.
The last car is unhooked from a moving train. The train continues to move at the same speed v 0 . How will the distances traversed by the train and the car be related to the moment the car stops? Assume that the car was moving at equal speed. Solve the problem also graphically.
Answer
At the moment when the train started, the person accompanying him began to run evenly along the train at a speed v 0 =3.5 m/s. Assuming the motion of the train to be uniformly accelerated, determine the speed of the train v at the moment when the person seeing off reaches the person seeing him off.
Answer
v=7 m/s.
A graph of the velocity of a certain body versus time is shown in the figure.
Draw graphs of the acceleration and coordinates of the body, as well as the distance traveled by it, versus time.
Answer
Graphs of the acceleration, coordinates of the body, as well as the distance traveled by it versus time are shown in the figure.
The graph of the acceleration of a body versus time has the form shown in the figure.
Draw graphs of speed, displacement and distance traveled by the body versus time. The initial velocity of the body is zero (at the discontinuity site the acceleration is zero).
The body begins to move from a point A with speed v 0 and after some time gets to the point B.
What distance did the body travel if it moved uniformly with an acceleration numerically equal to a? Distance between points A And B equals l. Find the average speed of the body.
The figure shows a graph of the body coordinates versus time.
After the moment t=t 1 curve of the graph is a parabola. What kind of movement is shown in this graph? Draw a graph of the body's speed versus time.
Solution
In the area from 0 to t 1: uniform movement with speed v 1 = tg α ;
in the area from t 1 to t 2: uniform slow motion;
in the area from t 2 to t 3: uniformly accelerated movement in the opposite direction.
The figure shows a graph of the velocity of a body versus time.
The figure shows velocity graphs for two points moving along the same straight line from the same initial position.
Known moments in time t 1 and t 2. At what point in time t Will 3 points meet? Construct motion graphs.
In what second from the beginning of the movement is the distance covered by the body in uniformly accelerated motion, three times the distance traveled in the previous second, if the movement occurs without an initial speed?
Answer and solution
In a second.
The easiest way to solve this problem is graphically. Because the path traveled by the body is numerically equal to the area of the figure under the line of the speed graph, then from the figure it is obvious that the path traveled in the second second (the area under the corresponding section of the graph is equal to the area of three triangles) is 3 times greater than the path traveled in the first second (the area is equal to the area one triangle).
The trolley must transport the cargo to the shortest possible time from one place to another at a distance L. It can accelerate or slow down its movement only with the same magnitude and constant acceleration a, then moving into uniform motion or stopping. What is the highest speed v must the trolley reach to fulfill the above requirement?
Answer and solution
It is obvious that the trolley will transport the cargo in the minimum time if it moves with acceleration for the first half of the journey + a, and the remaining half with acceleration - a.
Then we can write the following expressions: L = ½· vt 1 ; v = ½· at 1 ,
where we find the maximum speed:
Jet plane flies at speed v 0 =720 km/h. From a certain moment the plane moves with acceleration for t=10 s and v last second goes the way s=295 m. Determine acceleration a and final speed v airplane.
Answer and solution
a=10 m/s 2, v=300 m/s.
Let's draw a graph of the plane's speed in the figure.
Aircraft speed at time t 1 is equal v 1 = v 0 + a(t 1 - t 0). Then the distance traveled by the plane in the time from t 1 to t 2 is equal s = v 1 (t 2 - t 1) + a(t 2 - t 1)/2. From here we can express the desired acceleration value a and, substituting the values from the problem conditions ( t 1 - t 0 = 9 s; t 2 - t 1 = 1 s; v 0 = 200 m/s; s= 295 m), we get the acceleration a= 10 m/s 2. Final speed airplane v = v 2 = v 0 + a(t 2 - t 0) = 300 m/s.
The first car of the train passed the observer standing on the platform behind t 1 = 1 s, and the second - for t 2 =1.5 s. Car length l=12 m. Find acceleration a trains and their speed v 0 at the beginning of observation. The movement of the train is considered to be uniformly variable.
Answer and solution
a=3.2 m/s 2, v 0 ≈13.6 m/s.
The distance traveled by the train to the moment in time t 1 is equal to:
and the path to the moment in time t 1 + t 2:
.
From the first equation we find v 0:
.
Substituting the resulting expression into the second equation, we get the acceleration a:
.
A ball launched up an inclined plane passes two successively equal to the segment length l everyone continues to move on. The ball passed the first segment in t seconds, the second - in 3 t seconds Find speed v ball at the end of the first segment of the path.
Answer and solution
Since the motion of the ball under consideration is reversible, it is advisable to choose the starting point common point two segments. In this case, the acceleration when moving in the first segment will be positive, and when moving in the second segment - negative. The initial speed in both cases is equal v. Now let’s write down the system of equations of motion for the paths traversed by the ball:
Eliminating acceleration a, we get the required speed v:
A board divided into five equal segments begins to slide along an inclined plane. The first segment passed the mark made on the inclined plane in the place where the front edge of the board was at the beginning of the movement, beyond τ =2 s. For what time will pass past this mark is the last piece of the board? The movement of the board is considered to be uniformly accelerated.
Answer and solution
τ n =0.48 s.
Let's find the length of the first segment:
Now let's write down the equations of motion for the starting points (time t 1) and end (time point t 2) fifth segment:
By substituting the length of the first segment found above instead of l and finding the difference ( t 2 - t 1), we get the answer.
A bullet traveling at a speed of 400 m/s hits Earthworks and penetrates into it to a depth of 36 cm. How long did it move inside the shaft? At what acceleration? What was its speed at a depth of 18 cm? At what depth did the speed of the bullet decrease by a factor of three? The movement is considered uniformly variable. What will be the speed of the bullet by the time the bullet has traveled 99% of its path?
Answer and solution
t= 1.8·10 -3 s; a≈ 2.21·10 5 m/s 2 ; v≈ 282 m/s; s= 32 cm; v 1 = 40 m/s.
We will find the time of movement of the bullet inside the shaft from the formula h = vt/2, where h- full depth of immersion of the bullet, from where t = 2h/v. Acceleration a = v/t.
A ball was allowed to roll from bottom to top on an inclined board. On distance l= 30 cm from the beginning of the path the ball has visited twice: through t 1 = 1 s and after t 2 = 2 s after the start of movement. Define initial speed v 0 and acceleration a movement of the ball, considering it constant.
Answer and solution
v 0 = 0.45 m/s; a= 0.3 m/s 2.
The dependence of the ball speed on time is expressed by the formula v = v 0 - at. At a moment in time t = t 1 and t = t 2 the ball had the same velocities and opposite in direction: v 1 = - v 2. But v 1 =v 0 - at 1 and v 2 = v 0 - at 2, therefore
v 0 - at 1 = - v 0 + at 2, or 2 v 0 = a(t 1 + t 2).
Because the ball moves uniformly accelerated, then the distance l can be expressed as follows:
Now you can create a system of two equations:
,
solving which, we get:
A body falls from a height of 100 m without initial velocity. How long does it take for a body to pass through the first and last meters your way? How far does the body travel during the first and last second of its movement?
Answer
t 1 ≈ 0.45 s; t 2 ≈ 0.023 s; s 1 ≈ 4.9 m; s 2 ≈ 40 m.
Determine the time open position photographic shutter τ if, when photographing a ball falling along a vertical centimeter scale from the zero mark without an initial speed, a strip was obtained on the negative extending from n 1 to n 2 scale divisions?
Answer
.
A freely falling body traveled the last 30 m in a time of 0.5 s. Find the height of the fall.
Answer
A freely falling body has covered 1/3 of its path in the last second of its fall. Find the time of fall and the height from which the body fell.
Answer
t≈ 5.45 s; h≈ 145 m.
At what initial speed v 0 you need to throw the ball down from a height h so that he jumps to a height of 2 h? Neglect air friction and other losses of mechanical energy.
Answer
With what time interval did two drops come off the roof eaves, if two seconds after the second drop began to fall, the distance between the drops was 25 m? Neglect air friction.
Answer
τ ≈ 1 s.
The body is thrown vertically upward. An observer notices a period of time t 0 between two moments when the body passes the point B, located at a height h. Find the initial throwing speed v 0 and the time of the entire body movement t.
Answer
; 
.
From points A And B, located vertically (point A above) at a distance l= 100 m from each other, two bodies are thrown simultaneously with the same speed of 10 m/s: from A- vertically down, from B- vertically up. After how long and in what place will they meet?
Answer
t= 5 s; 75 m below point B.
A body is thrown vertically upward with an initial velocity v 0 . When it reached highest point path, from the same starting point at the same speed v 0 second body thrown. At what altitude h from the starting point will they meet?
Answer
Two bodies are thrown vertically upward from the same point with the same initial speed v 0 = 19.6 m/s with time interval τ = 0.5 s. After what time t after throwing the second body and at what height h will bodies meet?
Answer
t= 1.75 s; h≈ 19.3 m.
The balloon rises vertically upward from the Earth with acceleration a= 2 m/s 2. Through τ = 5 s from the beginning of its movement an object fell out of it. After how long t will this object fall to earth?
Answer
t≈ 3.4 s.
From a balloon descending at a speed u, throw the body up at speed v 0 relative to the Earth. What will be the distance l between the balloon and the body at the moment of the highest rise of the body relative to the Earth? What is the greatest distance l max between body and balloon? After what time τ from the moment of throwing the body will be level with the balloon?
Answer
l = v 0 2 + 2uv 0 /(2g);
l max = ( u + v 0) 2 /(2g);
τ = 2(v 0 + u)/g.
A body located at a point B on high H= 45 m from the Earth, begins to fall freely. Simultaneously from the point A, located at a distance h= 21 m below point B, throw another body vertically upward. Determine initial speed v 0 of the second body, if it is known that both bodies will fall to the Earth at the same time. Neglect air resistance. Accept g= 10 m/s 2.
Answer
v 0 = 7 m/s.
A body falls freely from a height h. At the same moment another body is thrown from a height H (H > h) vertically down. Both bodies fell to the ground at the same time. Determine initial speed v 0 second body. Check the correctness of the solution on numerical example: h= 10 m, H= 20 m. Accept g= 10 m/s 2.
Answer
v 0 ≈ 7 m/s.
A stone is thrown horizontally from the top of a mountain with a slope α. At what speed v 0 a stone must be thrown so that it falls on a mountain in the distance L from the top?
Answer
Two people play with a ball, throwing it to each other. What is the greatest height the ball reaches during a game if it flies from one player to another for 2 s?
Answer
h= 4.9 m.
The plane flies at a constant altitude h in a straight line at speed v. The pilot must drop the bomb on a target in front of the aircraft. At what angle to the vertical should he see the target at the moment the bomb is dropped? What is the distance from the target to the point over which the plane is located at this moment? Do not take into account air resistance to the movement of the bomb.
Answer
; .
Two bodies fall from the same height. On the path of one body there is a platform located at an angle of 45° to the horizon, from which this body is elastically reflected. How do the times and speeds at which these bodies fall differ?
Answer
The falling time of the body on the path of which the platform was located is longer, since the vector of the speed accumulated at the moment of impact changed its direction to horizontal (at elastic collision the direction of the velocity changes, but not its magnitude), which means the vertical component of the velocity vector became equal to zero, while the velocity vector of the other body did not change.
The falling velocities of the bodies are equal until the moment of collision of one of the bodies with the platform.
The elevator rises with an acceleration of 2 m/s 2 . At the moment when its speed became equal to 2.4 m/s, a bolt began to fall from the ceiling of the elevator. The height of the elevator is 2.47 m. Calculate the time the bolt falls and the distance traveled by the bolt relative to the shaft.
Answer
0.64 s; 0.52 m.
At a certain height, two bodies are simultaneously thrown from one point at an angle of 45° to the vertical with a speed of 20 m/s: one down, the other up. Determine height difference Δh, on which there will be bodies in 2 s. How do these bodies move relative to each other?
Answer
Δ h≈ 56.4 m; bodies move away from each other at a constant speed.
Prove that when free movement bodies near the Earth's surface, their relative speed is constant.
From point A the body falls freely. Simultaneously from the point B at an angle α another body is thrown towards the horizon so that both bodies collide in the air.
Show that the angle α does not depend on the initial speed v 0 body thrown from a point B, and determine this angle if . Neglect air resistance.
Answer
α = 60°.
Body thrown at an angle α towards the horizon at speed v 0 . Determine speed v this body is on top h above the horizon. Does this speed depend on the throwing angle? Ignore air resistance.
At an angle α =60° a body is thrown towards the horizon with an initial speed v=20 m/s. After how long t it will move at an angle β =45° to the horizon? There is no friction.
From three pipes located on the ground, jets of water shoot out at the same speed: at an angle of 60, 45 and 30° to the horizon. Find relationships highest altitudes h the rise of the jets of water flowing from each pipe, and the fall distances l water to the ground. Ignore air resistance to the movement of water jets.
From a point lying at the upper end of the vertical diameter d of a certain circle, along gutters installed along various chords of this circle, loads simultaneously begin to slide without friction.
Determine after what period of time t the loads will reach the circle. How does this time depend on the angle of inclination of the chord to the vertical?
Initial speed of a thrown stone v 0 =10 m/s, and after t=0.5 s stone speed v=7 m/s. On what maximum height above entry level will the stone rise?
Answer
H max ≈ 2.8 m.
At a certain height, simultaneously from one point with at the same speeds balls are thrown in all possible directions. What will be the geometric location of the points where the balls are located at any given time? Neglect air resistance.
Answer
Geometric place points of location of the balls at any moment of time there will be a sphere whose radius v 0 t, and its center is located below the starting point by an amount GT 2 /2.
The target located on the hill is visible from the gun location at an angle α to the horizon. The distance (horizontal distance from the gun to the target) is equal to L. Shooting at the target is carried out at an elevation angle β .
Determine initial speed v 0 projectile hitting the target. Ignore air resistance. At what elevation angle β 0 will the firing range along the slope be maximum?
Answer and solution
, .
Let's choose a coordinate system xOy so that the reference point coincides with the tool. Now let’s write down the kinematic equations of projectile motion:
Replacing x And y to the target coordinates ( x = L, y = L tgα) and excluding t, we get:
Range l projectile flight along a slope l = L/cos α . Therefore, the formula we received can be rewritten as follows:
.
,
this expression is maximum at maximum value works
That's why l maximum with maximum value = 1 or
At α = 0 we get the answer β 0 = π /4 = 45°.
An elastic body falls from a height h on inclined plane. Determine how long t after reflection, the body will fall onto an inclined plane. How does time depend on the angle of the inclined plane?
Answer
Does not depend on the angle of the inclined plane.
From high H on an inclined plane forming an angle with the horizon α =45°, the ball falls freely and is elastically reflected at the same speed. Find the distance from the place of the first impact to the second, then from the second to the third, etc. Solve the problem in general view(for any angle α ).
Answer
; s 1 = 8H sin α ; s 1:s 2:s 3 = 1:2:3.
The distance to the mountain is determined by the time between the shot and its echo. What can be the error τ in determining the moments of a shot and the arrival of an echo, if the distance to the mountain is at least 1 km, and it needs to be determined with an accuracy of 3%? Speed of sound in air c=330 m/s.
Answer
τ ≤ 0.09 s.
They want to measure the depth of the well with an accuracy of 5% by throwing a stone and noting the time τ , through which the splash will be heard. Starting from what values τ Is it necessary to take into account the sound travel time? Speed of sound in air c=330 m/s.
Answer
Problem 1. The minimum time required to cross a river in a boat is t o. The width of the river bed is H. The speed of the river flow is constant anywhere in the channel u V β times the speed of the boat ( β > 1), floating in still water.
- Find the speed of the boat in still water.
- How far will the boat move in the minimum crossing time?
- Determine the shortest distance that the boat can drift during the crossing.
- Find the time it takes for the boat to cross when it is drifted a minimum distance.
1.
The minimum distance between banks is the width of the river. If you point the boat perpendicular to the shore, then its movement time will be minimal t = H/v o, because H− minimal, and v L− is maximum, then
v L = H/t o. (1)
2.
Since the speed vector of the boat is directed perpendicular to the shore, the drift of the boat depends only on the speed of the current. River flow speed v T = βv L; During the crossing the boat will be carried away
L = v T t o = βv L t o = βHt o /t o = βH.
The demolition of the boat (in the minimum time of movement) will be
L = βH. (2)
3.
The drift of the boat during the crossing will depend on two factors: the speed of the boat in the direction of the current and the speed of the boat in the direction perpendicular to the shore. It is necessary to determine the angle of the boat's speed vector. Relatively in a simple way finding the angle is graphic method. The speed of the boat relative to the coordinate system associated with the shore is equal to vector sum speeds of the current and the boat (Fig.). The figure shows that the minimum distance Lmin boat drift corresponds to the case when the relative speed of the boat is directed tangentially to a circle of radius v L. From the similarity of triangles of speeds and distances having common angle α
, we get
L min /H = v/v L,
and since v ⊥ v o, we find
L min = Hv/v Л = H√(v T 2 − v Л 2 ) = H√(β 2 (H/t o) 2 − (H/t o) 2 ) = H√(β 2 − 1). (3)
4.
The time it takes for a boat to cross when it drifts a minimum distance depends on the projection of the boat's speed onto the axis Oy.
Projection of boat speed on Oy equal to
v y = v Л cosα.
On the other side
.
Transit time in this case
t = Hβ/(v L √(β 2 − 1)) = βt o /√(β 2 − 1). (4)
Note 1. The minimum time for a boat to cross the river will be if the boat moves perpendicular to the shore.
Note 2. The minimum drift of the boat will be in the case when the speed vector of the boat is perpendicular to the vector relative speed boats.
Note 3. Determining the angle between the speed vector of the boat and (for example) the vertical, for minimal drift when crossing a river, is possible in the following ways:
Through the study of function. When crossing to the other side
H = v L cosα × t And L = (v T − v Л sinα)t.
Let's create a trajectory equation L(H)
L = (v T − v Л sinα)H/(v Л cosα) = v T H/(v Л cosα) − Htgα.
Finally, L = v T H/(v Л cosα) − Htgα.
Differentiating the last equation with respect to the angle α
and, equating the derivative to zero, we find at what values of the angle α
distance L will be minimal.
(v T H/(v L cosα) − Htgα) / = v T Hsinα/(v L cos 2 α − H/cos 2 α), sinα = v L /v T = 1/β.
Through the trigonometric unit
sin 2 α + cos 2 α = 1, we'll find cosα = √(β 2 − 1)/β.
Discriminant method. We rewrite the trajectory equation in the form
L = v T H/(v Л cosα − Hsinα/cosα)
or
Lcosα = βH − Hsinα.
Let's square the equation
L 2 cos 2 α = β 2 H 2 + H 2 sin 2 α − 2βH 2 sinα.
Using the trigonometric unit
sin 2 α + cos 2 α = 1.
Then
L 2 (1 − sin 2 α) = β 2 H 2 + H 2 sin 2 α − 2βH 2 sinα.
We have obtained a quadratic equation for the desired angle α
. Let's transform it to a “normal” (convenient form).
(L 2 + H 2)sin 2 α − 2βH 2 sinα − (L 2 − (βH) 2) = 0.
Solution quadratic equation has the form:
sinα 1,2 = (βH 2 ± √((βH 2) 2) − (β 2 H 2 − L 2)(L 2 + H 2)))/(L 2 + H 2).
Wherein D ≥ 0:
β 2 H 4) − (β 2 H 2 − L 2)(L 2 + H 2) = L 2 (L 2 − β 2 H 2 + H 2) ≥ 0.
When decreasing L the discriminant decreases. Minimum value D=0. Then,
L 2 = β 2 H 2 − H 2 , and L = H√(β 2 − 1),
which corresponds to minimal drift.
From the figure it is clear that
cosα = L min /√(L min 2 + H 2 ) = H√(β 2 − 1)/√(H 2 (β 2 − 1) + H 2 ) = √(β 2 − 1)/β.
Note 4. If the current speed is less than the speed of the boat, then minimal drift is possible only when the boat moves in a minimum time (see solution 1).
Problems to solve independently.
1.
The boat, crossing a river 800 m wide, moved at a speed of 4 m/s so that its crossing time turned out to be minimal. How much will the boat be carried away by the current if the speed of the river is 1.5 m/s?
2. When crossing a river 60 m wide, you need to get to a point located 80 m downstream than the starting point. The boatman controls the motor boat so that it moves exactly towards the target at a speed of 8 m/s relative to the shore. What is the speed of the boat relative to the water if the speed of the river is 2.8 m/s?
3. At what angle to the shore should a motor boat go in order to cross a river 300 m wide in the minimum time, if the speed of the boat relative to the water is 18 km/h and the speed of the current is 2 m/s? How much will the boat move along the shore?
4. A boat is crossing a river, starting from point A. The speed of the boat in still water is 5 m/s, the speed of the river current is 3 m/s, the width of the river is 200 m. a) At what point will the boat land on the opposite bank if the course is perpendicular to the banks? b) What course should be taken to get to point B, located on the opposite bank opposite point A? For both cases, find the crossing time.
5. A swimmer wants to cross a river of width h. At what angle α to the direction of the river flow should he swim to cross least time? Which way will he swim? The speed of the river current is u, the speed of the swimmer relative to the water is v. How long will it take him to swim across the river? the shortest path? [α = 90°; l = h√(u 2 + v 2)/v]
6. Two boats left simultaneously from points A and B located on different banks, with point B downstream. Both boats are moving along a straight line AB, the length of which is l = 1 km. Straight AB forms an angle α = 60° with the direction of the flow velocity, which is equal to v = 2 m/s. The boats met 3 minutes after leaving the berths. At what distance from point B did the meeting take place?
7. A tourist kayaking down the river noticed that the stream was carrying him to the middle of a tree that had fallen and blocked his path at the moment when the distance from the bow of the kayak to the tree was S = 30 m. Estimate at what angle to the speed of the current he should steer kayak to get around the obstacle. The speed of the river is u = 3 km/h, the speed of the kayak relative to the water is 6 km/h, the length of the tree is l = 20 m. [α = 31°]
8. The speed of the river current is 5 m/s, its width is 32 m. Crossing the river in a boat whose speed relative to the water is 4 m/s, the helmsman ensured the smallest possible drift of the boat by the current. What is this demolition equal to?
9. From point A, located on the river bank, you need to get to point B, located on the opposite bank, upstream at a distance of 2 km from the perpendicular drawn from point A to the opposite bank. River width 1 km, maximum speed the boat relative to the water is 5 km/h, and the speed of the river is 2 km/h. Will the boat be able to swim across to the other side in 30 minutes, moving along straight AB.
10. Two motor boats located opposite each other on opposite banks straight section with a width H = 200 m, they cross in such a way that the time it takes one boat to cross and the movement of the other boat during its crossing is minimal. The speed v = 5 m/s of each boat relative to the water is n = 2 times the speed of the current. Find the minimum distance between the boats and the time T for their movement to move closer to this distance if the boats begin crossing at the same time. The speed of the current and the speed of movement of each boat during the crossing are considered constant.
See also:Reminder for completing tasks:
· Read the problem statement carefully;
· Repeat the conditions of the problem and questions;
· Think about what is known and what needs to be found;
· Analyze the solution to the problem: what needs to be found at the beginning and what at the end;
· Make a plan for solving the problem, solve the problem;
· Check the progress of the solution, the answer.
The solution and answers are included in Text Document located below. Don't forget to include your full name and task number.Problem No. 3 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade.Problem condition: it is known that the mass of the Sun is 330,000 times more mass Earth. Is it true that the Sun attracts the Earth 330,000 times stronger than the Earth attracts the Sun? Explain your answer.Problem No. 4 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
The boat has moved relative to the pier from point A(-8; -2) to point B(4; 3). Make a drawing, aligning the origin with the pier and indicating points A and B on it. Determine the displacement of the boat AB. Could the distance traveled by the boat be greater than the movement it made? less movement? equal to displacement? Justify all answers.
Problem No. 5 from the textbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task: It is known that to determine the coordinates of a rectilinearly moving body, the equation x = x0 + sx is used. Prove that the coordinate of the body when it is rectilinear uniform motion for any moment in time is determined using the equation x = x0 + vxt
Problem No. 6 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
Write down an equation to determine the coordinates of a body moving rectilinearly at a speed of 5 m/s along the X axis, if at the moment the observation began its coordinate was 3 m.
Problem No. 7 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
Two trains - passenger and freight - move along parallel paths. Relative to the station building, the movement of a passenger train is described by the equation x p = 260 - 10t, and that of a freight train by the equation x t = -100 + 8t. Mistaking the station and trains for material points, indicate on the X-axis their positions at the moment the observation began. After what period of time from the beginning of observation did the trains meet? What is the coordinate of their meeting place? Specify the location of the meeting point on the X axis. Assume that the X axis is parallel to the rails.
Problem No. 9 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
A boy slides down a mountain on a sled, moving from a state of rest in a straight line and uniformly accelerated. In the first 2 s after the start of movement, its speed increases to 3 m/s. After what period of time from the beginning of the movement will the boy’s speed become equal to 4.5 m/s? How far will he travel during this period of time?
Problem No. 13 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
Two elevators - a regular one and a high-speed one - start moving at the same time and move at uniform acceleration for the same period of time. How many times is the distance covered by a high-speed elevator during this time greater than the distance covered by a regular elevator, if its acceleration is 3 times higher than the acceleration of a regular elevator? How many times greater speed will the high-speed elevator acquire by the end of this period of time compared to a regular elevator?
Problem No. 16 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task: When hit with a stick, the puck acquired an initial speed of 5 m/s and began to slide across the ice with an acceleration of 1 m/s2. Write down the equation for the dependence of the projection of the puck's velocity vector on time and construct a graph corresponding to this equation.
Problem No. 18 from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task: A skier rolls down a mountain, moving in a straight line with a constant acceleration of 0.1 m/s2. Write equations expressing the time dependence of the coordinates and projection of the skier’s speed vector, if his initial coordinates and speed are zero.
Problem No. from the workbook "Physics. 9th grade" by A.V. Peryshkin for 9th grade. The task:
A cyclist moves along a highway in a straight line at a speed whose modulus is 40 km/h relative to the ground. A car is moving parallel to it. What can be said about the magnitude of the velocity vector and the direction of movement of the car relative to the ground, if relative to the cyclist the magnitude of its (car) speed is equal to: a) 0; b) 10 km/h; c) 40 km/h; d) 60 km/h?
Some problems consider the motion of a body relative to another body, which is also moving in a chosen reference frame. Let's look at an example.
A raft is floating along the river, and a person is walking along the raft in the direction of the river flow - in the direction where the raft is floating (Fig. 3.1, a). Using a pole installed on the raft, it is possible to mark both the movement of the raft relative to the shore and the movement of a person relative to the raft.
Let us denote the speed of a person relative to the raft by pb, and the speed of the raft relative to the shore by pb. (It is usually assumed that the speed of the raft relative to the shore is equal to the speed of the river flow. We will denote the speed and movement of body 1 relative to body 2 using two indices: the first index refers to body 1, and the second to body 2. For example, 12 denotes the speed of the body 1 relative to body 2.) 
Let us consider the movements of a person and a raft over a certain period of time t.
Let us denote by PB the movement of the raft relative to the shore, and by pm - the movement of a person relative to the raft (Fig. 3.1, b).
Displacement vectors are shown in the figures with dotted arrows to distinguish them from velocity vectors, shown with solid arrows.
The movement of a bw person relative to the shore is equal to the vector sum of the person’s movement relative to the raft and the movement of the raft relative to the shore (Fig. 3.1, c):
Bw = pb + bp (1)
Let us now connect the movements with speeds and time interval t. We will get:
Chp = chp t, (2)
pb = pb t, (3)
bw = bw t, (4)
where bw is the speed of a person relative to the shore.
Substituting formulas (2–4) into formula (1), we obtain:
Bw t = pb t + bp t.
Let's reduce both sides of this equation by t and get:
Bw = pb + chp. (5)
Speed addition rule
Relation (5) is the rule for adding velocities. It is a consequence of the addition of displacements (see Fig. 3.1, c, below). In general, the rule for adding speeds looks like this:
1 = 12 + 2 . (6)
where 1 and 2 are the velocities of bodies 1 and 2 in the same reference frame, and 12 is the speed of body 1 relative to body 2.
So, the speed 1 of body 1 in a given frame of reference is equal to the vector sum of the speed 12 of body 1 relative to body 2 and the speed 2 of body 2 in the same frame of reference.
In the example discussed above, the speed of the person relative to the raft and the speed of the raft relative to the shore were in the same direction. Now consider the case when they are directed in the opposite direction. Do not forget that the velocities must be added according to the rule of vector addition!
1. The man is walking along the raft against the current (Fig. 3.2). Make a drawing in your notebook that can be used to find the speed of a person relative to the shore. Scale for the velocity vector: two cells correspond to 1 m/s.
It is necessary to be able to add speeds when solving problems that involve the movement of boats or ships on a river or the flight of an airplane in the presence of wind. Wherein flowing water or moving air can be thought of as a “raft” that moves at a constant speed relative to the ground, “carrying” ships, airplanes, etc.
For example, the speed of a boat floating on a river relative to the shore is equal to the vector sum of the speed of the boat relative to the water and the speed of the river current.
2. The speed of a motor boat relative to the water is 8 km/h, and the speed of the current is 4 km/h. How long will it take the boat to travel from pier A to pier B and back if the distance between them is 12 km?
3. A raft and a motor boat set sail from pier A at the same time. During the time it took the boat to reach pier B, the raft had covered a third of this distance.
b) How many times is the time it takes the boat to move from B to A than the time it takes to move from A to B?
4. The plane flew from city M to city N in 1.5 hours at tailwind. The return flight with a headwind took 1 hour 50 minutes. The aircraft's speed relative to the air and the wind speed remained constant.
a) How many times is the speed of the airplane relative to the air greater than the speed of the wind?
b) How long would it take to fly from M to N in calm weather?
2. Transition to another reference system
It is much easier to track the motion of two bodies if you switch to the frame of reference associated with one of these bodies. The body with which the reference frame is connected is at rest relative to it, so you only need to monitor the other body.
Let's look at examples.
A motor boat overtakes a raft floating on the river. An hour later, she turns around and swims back. The speed of the boat relative to the water is 8 km/h, the speed of the current is 2 km/h. How long after the turn does the boat meet the raft?
If we solve this problem in a reference frame associated with the shore, then we would have to monitor the movement of two bodies - the raft and the boat, and also take into account that the speed of the boat relative to the shore depends on the speed of the current.
If we go to the frame of reference associated with the raft, then the raft and the river will “stop”: after all, the raft moves along the river exactly at the speed of the current. Therefore, in this reference frame, everything happens as in a lake where there is no current: the boat floats from the raft and to the raft with the same absolute speed! And since she moved away for an hour, then in an hour she will sail back.
As you can see, neither the speed of the current nor the speed of the boat were needed to solve the problem.
5. While passing under a bridge on a boat, a man dropped into the water straw hat. Half an hour later, he discovered the loss, swam back and found a floating hat at a distance of 1 km from the bridge. At first, the boat floated with the current and its speed relative to the water was 6 km/h.
Go to the frame of reference associated with the hat (Figure 3.3) and answer the following questions.
a) How long did the man swim to the hat?
b) What is the speed of the current?
c) What information in the condition is not needed to answer these questions?
6. A foot column 200 m long is walking along a straight road at a speed of 1 m/s. The commander at the head of the column sends a rider with an order to the trailing one. How long will it take the rider to get back if he gallops at a speed of 9 m/s?
Let's deduce general formula to find the speed of a body in a reference frame associated with another body. To do this, we will use the rule of adding velocities.
Recall that it is expressed by the formula
1 = 2 + 12 , (7)
where 12 is the speed of body 1 relative to body 2.
Let us rewrite formula (1) in the form
12 = 1 – 2 , (8)
where 12 is the speed of body 1 in the reference frame associated with body 2.
This formula allows you to find the speed 12 of body 1 relative to body 2 if the speed 1 of body 1 and the speed 2 of body 2 are known.
7. Figure 3.4 shows three cars, the speeds of which are given on a scale: two cells correspond to a speed of 10 m/s. 
Find:
a) the speed of the blue and purple cars in the reference frame associated with the red car;
b) the speed of the blue and red cars in the frame of reference associated with the purple car;
c) the speed of the red and purple cars in the reference frame associated with the blue car;
d) which of the found speeds is the greatest in absolute value? smallest?
Additional questions and tasks
8. A man walked along a raft of length b and returned to the starting point. The speed of a person relative to the raft is always directed along the river and is equal in magnitude to vh, and the speed of the current is equal to vt. Find an expression for the path traveled by a person relative to the shore if:
a) at first the person walked in the direction of the current;
b) at first the person walked in the opposite direction to the flow (consider all possible cases!).
c) Find the entire path traveled by the person relative to the shore: 1) at b = 30 m, v h = 1.5 m/s, v t = 1 m/s; 2) at b = 30 m, v h = 0.5 m/s, v t = 1 m/s.
9. A passenger on a moving train noticed that two oncoming trains rushed past his window with an interval of 6 minutes. At what interval did they pass the station? The speed of the train is 100 km/h, the speed of electric trains is 60 km/h.
10. Two people simultaneously started descending the escalator. The first one stood on one step. At what speed did the second person walk down the escalator if he went down 3 times faster than the first one? Escalator speed 0.5 m/s.
11. The escalator has 100 steps. A man walking down an escalator counted 80 steps. How many times is the speed of a person greater than the speed of an escalator?
12. A raft and a motor boat departed from pier A at the same time. While the raft reached pier B, the boat swam from A to B and back. The distance AB is 10 km.
a) How many times is the speed of the boat relative to the water greater than the speed of the current?
b) How far did the raft travel when: 1) the boat reached B? 2) did the raft meet a boat sailing back?
13. The fastest animal is the cheetah (Fig. 3.5): it can rush at a speed of 30 m/s, but no more than one minute. The cheetah noticed an antelope located at a distance of 500 m from him. How fast should the antelope run to escape? 