Kinematics

1. From points and , separated by a distance , two bodies start moving simultaneously towards each other: the first with velocity , the second with . Determine the time until they meet and the distance from point to the meeting point. Solve the problem graphically as well.

Answer

; .

Solution

The dependence of the coordinates of the bodies on time is given by the formulas: The meeting () will occur after time from the start of motion at a point located at a distance from point A. The graphs of the dependence of the coordinates of the bodies on time are shown below (; , since is directed in the opposite direction to the one taken as positive). The moment of meeting corresponds to point C of intersection of the graphs.
Description

2. After how much time and where would the bodies (see See Problem 1) meet if they moved in the same direction , with the body from point starting to move seconds after the body from point ?

Answer

; .

Hint

Use the graphs of the dependence of the coordinates of the bodies on time (the following figure). Description

3. A motorboat covers the distance between two points and downstream in time , while a raft takes . How much time will the motorboat take for the return journey?

Answer

Solution

Let be the speed of the current, and be the speed of the boat relative to the water, we can write: To determine the time taken by the boat for the return trip, it is convenient to rewrite equations (1) and (2) as follows: Subtracting twice equation (4) from equation (3), we get whence

4. An escalator in the metro descends a person walking down it in 1 minute. If the person walks twice as fast, they descend in 45 seconds. How much time does it take for a person standing still on the escalator to descend?

Answer

1.5 min.

5. A person runs on an escalator. The first time, they count steps; the second time, moving in the same direction at three times the speed, they count steps. How many steps would they count on a stationary escalator?

Answer

Solution

If the person’s velocity were directed opposite to the motion of the escalator, then the faster he walked, the fewer steps he would count (but no fewer than . In our case, the directions of the person’s and the escalator’s motion coincide.

Let be the speed of the escalator; its length, and the number of steps on a stationary escalator. The number of steps per unit length of the escalator is .

Therefore, if the person walks with speed relative to the escalator, the time spent on the escalator is , and the distance he travels (relative to the escalator) is .

Thus, the number of steps the person counts is:

Similarly, in the second case (if the walking speed is tripled), he counts:

Hence, we obtain the following system of equations:

From this, by eliminating the ratio , we find:

6. Between two points on a river, located apart, a boat shuttles. Going downstream, it covers this distance in , and upstream in . Determine the river's current speed and the boat's speed relative to the water.

Answer

km/h; km/h.

7. A raft passes by a dock. At that moment, a motorboat departs downstream to a village located from the dock. The boat reaches the village in and, turning back, meets the raft at a distance from the village. What are the river's current speed and the boat's speed relative to the water?

Answer

km/h; km/h.

Solution

Let's choose the raft (water) as the reference frame. In this reference frame, the boat moves down and up the river with the same speed. This means that the time for the boat to move away from the raft is equal to the time to approach it; thus, the boat also returned in 3/4 h. During the elapsed 1.5 h, the raft traveled a distance Consequently, the speed of the current (speed of the raft relative to the bank) is The speed of the boat relative to the water

This solution illustrates how important a successful choice of reference frame is in kinematics.

8. A column of troops marches at a speed of , stretched along a road for a distance . A commander at the rear sends a cyclist with a message to the lead detachment. The cyclist rides at , delivers the message, and immediately returns at the same speed. How much time after receiving the message does the cyclist return?

Answer

min.

Hint

The speed of the cyclist in the reference frame associated with the column is when moving towards the head of the column, and when returning. Therefore

9. A train car of width , moving at , is pierced by a bullet flying perpendicular to the car's motion. The displacement of the holes in the car's walls relative to each other is . What is the bullet's speed?

Answer

600 m/s.

10. What is the speed of vertically falling raindrops if a car driver notices that the drops leave no trace on the rear window, tilted forward at an angle to the horizon, when the car's speed exceeds 30 km/h?

Answer

m/s.

Solution

For the raindrops not to leave a trace on the rear window, the direction of the raindrop velocity vector relative to the car must make an angle with the horizon no greater than . The relative velocity is composed of the vertical component, equal to the falling speed of the drop , and the horizontal component, equal to the speed of the car and directed in the opposite direction (following figure). From the velocity triangle, it is clear that .
Description

11. It is raining outside. In which case will a bucket standing in the back of a truck fill faster with water: when the truck is moving or when it is stationary?

Answer

Identical.

12. At what speed and course must an airplane fly to cover a path of due North in time , if a northwest wind blows at an angle to the meridian with speed during the flight?

Answer

km/h; northwest at an angle to the meridian.

Hint

Let be the speed of the aircraft relative to the Earth. It is directed, according to the condition, along the meridian; then . By the cosine theorem The angle between and is found from the sine theorem:

13. A blackboard moves on a smooth horizontal table at speed . What shape will chalk leave as a trace on the board if it is thrown horizontally at speed perpendicular to the board's motion, assuming: a) negligible friction between the chalk and the board; b) significant friction?

Answer

a) and b) - a straight line forming an angle with the direction of motion of the board. In case b), the trace may not reach the edge of the board.

Hint

Consider the motion of the chalk in the coordinate system associated with the board. Since the friction force is directed along the velocity vector of the chalk relative to the board, it cannot change the direction of the chalk's motion and only reduces its speed.

14. A ship departs from point at speed , making an angle with the line (following figure). At what angle to the line should a torpedo be launched from point to hit the ship? The torpedo must be launched when the ship is at point . The torpedo's speed is . Description

Answer

Hint

Point C (see figure below) is the meeting place of the ship and the torpedo. , , where is the travel time of the torpedo. According to the sine theorem or Hence Description

15. A cord is attached to a slider that can move along a guide rail (see figure below) and threaded through a ring. The cord is pulled at speed . At what speed does the slider move when the cord makes an angle with the guide rail? Description

Answer

Solution

For the same small time interval , the slider moves by , and the cord is pulled by the length (see figure) ( can be considered right due to the smallness of ). Therefore, we can write , whence , i.e., the speed of pulling the rope is equal to the projection of the slider's velocity onto the direction of the rope.
Description

16. Workers lifting a load (following figure) pull ropes at the same speed . What speed does the load have when the angle between the ropes to which it is attached is ? Description

Answer

Hint

The projection of the velocity of the load onto the direction of each cable must be equal to the speed of the cable (see problem 15).

17. A rod of length is connected by hinges to collars and , which move along two mutually perpendicular rails (see figure). Collar moves at a constant speed . Find the speed of collar when the angle . Taking the time origin as when collar was at point , determine the distance and the speed of collar as a function of time. Description

Answer

cm/s; ; .

Hint

See solution to problem 15. At any moment in time, the projections of the velocities and of the ends of the rod (see figure below) onto the axis of the rod are equal to each other, otherwise the rod would have to shorten or lengthen. Thus, , whence . Description

18. A tank moves at . At what speed do the following move relative to the Earth: a) the upper part of the track;

b) the lower part of the track;

c) the point on the track moving vertically relative to the tank at that moment?

Answer

a) 40 m/s;

b) 0;

c) m/s.

Hint

The velocity of any point on the track relative to the Earth at any given moment is composed of the velocity of the tank relative to the Earth and the velocity of the track link relative to the tank.

19. 1. A car travels the first half of the path at and the second half at . Find the average speed over the entire path. 2. A car travels half the path at , then half the remaining time at , and the last segment at . Find the car's average speed over the entire path.

Answer

1. km/h;

km/h.

Solution

The average speed is numerically equal to the ratio of the total path traveled by the body to the time of motion: Time therefore

2.Hint. The car traveled the second half of the path with an average speed equal to .

20. A train travels the first half of the path at a speed times greater than the second half. Its average speed over the entire path is . What are the train's speeds and for the first and second halves of the path?

Answer

km/h; km/h.

21. Two balls start moving simultaneously with the same speed along surfaces shaped as shown in the following figure. How will their speeds and travel times differ when they arrive at point ? Neglect friction. Description

Answer

The speeds will be the same. The travel time of the second ball is less.

Solution

Sample graphs of the speed of the balls are shown in the following figure. Since the paths traveled by the balls are equal, as seen from the graph (on the graph, the path is numerically equal to the areas of the shaded figures), . Description

22. An airplane flies from point to point and back. Its speed in calm weather is . Find the ratio of the average speeds for the entire flight in two cases: a) when the wind blows along the line ; b) when the wind blows perpendicular to . The wind speed is .

Answer

Solution

If the wind blows along the line AB, the total flight time is and the average speed where is the distance between cities. If the wind is perpendicular to the line AB, the speed of the aircraft relative to the wind must be directed at an angle to the line AB, such that the drift is compensated (see figure). The flight speed in this case is constant and (the flight time in this case is ).
Description

23. A metro train covers the distance between two stations at an average speed . It spends accelerating, then moves uniformly for time , and takes decelerating to a full stop. Plot the train's speed graph and determine its maximum speed .

Answer

m/s. The graph of the train's speed is shown in the following figure.
Description

24. The last car detaches from a moving train. The train continues at the same speed . What is the ratio of the distances traveled by the train and the car when the car stops? Assume the car decelerates uniformly. Solve the problem graphically as well.

Answer

2:1; see the following figure.

Hint

The path traveled by the train after the car was detached is equal to the area of the rectangle OABC; the path traveled by the car is the area of the triangle OAC. Description

25. When the train starts moving, a well-wisher begins running uniformly alongside it at . Assuming the train's motion is uniformly accelerated, determine its speed when the well-wisher catches up.

Answer

m/s; see the following figure.

Hint

From the graph, it is clear that the paths traveled by the passerby and the train are equal at the moment when , where is the speed of the passerby. Description

26. The graph of a body's velocity versus time is shown in the following figure. Plot the graphs of acceleration, position, and distance traveled versus time. Description

Answer

Figure below.

Hint

The tangent of the angle of inclination of the tangent to the graph of the dependence of the coordinate (or path) of the body on time is numerically equal to the speed of the body. Therefore, the straight line representing the graph between and at the points corresponding to moments and will be tangent to the parabolas describing the motion of the body before moment and after moment , which are the graphs or before moment and after moment respectively. At moments and , the graphs and have a horizontal tangent. Description

27. The graph of a body's acceleration versus time is shown in the following figure. Plot the graphs of velocity, displacement, and distance traveled versus time. The initial velocity is zero (acceleration is zero at the discontinuity). Description

Answer

See the following figure.

Hint

The straight line between points and is tangent to the curves and describing the motion of the body before moment and after moment . Description

28. A body starts moving from point at speed and reaches point (see figure) after some time. What distance did the body travel if it moved with uniform acceleration ? The distance between and is . Find the body's average speed. Description

Answer

; .

Solution

Let's introduce a coordinate system with the origin at point A and the x-axis directed along the velocity of the body at point A (see figure). When the body reaches point B, its coordinate will be . The body reaches point B only if its acceleration is directed opposite to the velocity . Therefore, the dependence of velocity on time is expressed by the formula . At the moment , the velocity becomes zero. At this moment, the body is at the maximum distance from point A. For , . Substituting into the kinematic equation of motion: , we find . The path traveled by the body during the entire movement is To find the average speed , you need to know the time of motion of the body from point A to point B. It can be found by substituting the coordinate value of point B into the equation for x, i.e., : (the second root is negative, so it must be discarded). Then the average speed of the body is Description

29. The following figure shows the graph of a body's position versus time. After , the curve is a parabola. What motion is depicted? Plot the velocity versus time graph. Description

Answer

See the following; . Point corresponds to the maximum coordinate of the body. Before moment , the motion is uniform, for - uniformly accelerated with negative acceleration (i.e., acceleration directed opposite to velocity ). Description

30. The following figure shows velocity graphs for two points moving along the same line from the same starting position. The times and are known. At what time will the points meet? Plot the motion graphs. Description

Answer

Solution

Equating the paths traveled by the bodies by the moment of meeting: From the above figure, it is clear that (; ). Substituting the ratio into equation (1) and solving the resulting equation for , we find . The second root is discarded, as the condition must be met. The graphs of motion are shown in the following.
Description

31. In uniformly accelerated motion starting from rest, during which second does the distance traveled become three times the distance traveled in the previous second?

Answer

During the second second.

32. A cart must transport cargo the shortest possible distance . It can only accelerate or decelerate with constant magnitude , then move uniformly or stop. What maximum speed must the cart reach to meet this requirement?

Answer

Solution

The time to transport the cargo will be minimum if the average speed of movement of the trolley is maximum, which is obvious. The latter, under the conditions of this problem (start and end of motion occur with constant acceleration), can only be in the case where the trolley moves the first half of the path with acceleration , and the second - with acceleration (see figure). Thus, we can write the following relations: , , whence .
Description

33. A jet plane flies at . At some point, it accelerates for , covering in the last second. Determine the acceleration and final speed .

Answer

m/s; m/s.

34. The first train car passes an observer on the platform in , and the second in . Each car is long. Find the train's acceleration and initial speed . Assume uniformly accelerated motion.

Answer

m/s; m/s.

Hint

The paths traveled by the train by moments and are respectively equal: and .

35. A ball rolled up an inclined plane covers two equal segments of length each and continues further. The first segment takes seconds, the second seconds. Find the ball's speed at the end of the first segment.

Answer

Hint

The problem is solved most simply if we use the reversibility of motion and choose the boundary between the two segments as the origin of coordinates. Then the ball has an initial velocity - the sought value , while the acceleration in the direction towards the initial point is , and towards the final point is . Writing the equation of motion for the paths traveled by the ball, ; , and eliminating , we find the answer.

36. A board divided into five equal segments starts sliding down an inclined plane. The first segment passes a mark on the plane in . How long will it take for the last segment to pass the mark? Assume uniformly accelerated motion.

Answer

Solution

Let the entire length of the board be , then we can write that the time to pass the first segment is . The times to pass the beginning and end of the n-th segment past the observer are respectively equal: and whence

37. A bullet flying at strikes an earthen embankment and penetrates to a depth of . How long did it move inside the embankment? What was its acceleration? What was its speed at a depth of ? At what depth did its speed reduce by a factor of three? Assume uniformly decelerated motion. What will its speed be when it has traveled 99% of its path?

Answer

s; m/s; m/s; cm; m/s.

Hint

When determining at what depth the bullet's speed decreased by three times, use the formula . Starting from the point where the speed decreased by three times, the remaining path will be times less than the total path.

38. A ball rolled up an inclined board passes a point from the start twice: at and after starting. Determine the ball's initial speed and acceleration , assuming it is constant.

Answer

cm/s; cm/s.

Solution

1st method. The dependence of the coordinate of the body along the inclined plane on time is expressed by the formula . Hence Since and are the roots of this equation for , then, according to Vieta's theorem, From the obtained system of equations, it is easy to find and .

2nd method. The dependence of the ball's speed on time is expressed by the formula . At time moments and , the ball had speeds equal in magnitude and opposite in direction: . But therefore

Since the ball moves with uniform acceleration, its average speed over time is Therefore,
Solving equations (1) and (2) together, we find

39. A body falls from a height of without initial speed. How long does it take to cover the first and last meters of its path? What distance does it cover in the first and last seconds of its motion?

Answer

s; s; m; m.

40. Determine the open time of a photographic shutter if, when photographing a ball falling vertically along a centimeter scale starting from zero with no initial speed, the negative shows a streak extending from to scale divisions.

Answer

41. A freely falling body covers the last in . Find the height of the fall.

Answer

42. A freely falling body covers of its path in the last second. Find the fall time and height.

Answer

s; m.

Hint

Let be the height from which the body fell, and be the time of its fall from this height, we can write the following equations: ; . Solving them together, we get and . The problem can be solved more shortly if we use the formula for the path traveled by the body in the t-th second:

43. With what initial speed must a ball be thrown downward from height to bounce up to height ? Neglect air resistance and energy losses.

Answer

44. Two drops fall from a roof eaves. Two seconds after the second drop starts falling, the distance between them is . What is the time interval between their departures? Neglect air resistance.

Answer

Hint

Let: be the time interval between the detachment of the 1st and 2nd drops, be the time from the moment of detachment of the 2nd drop, be the distance between the drops after time (from the moment of detachment of the 2nd drop) and considering that the drops move relative to each other uniformly, we can write the equation to determine :

45. A body is thrown vertically upward. An observer notes a time interval between two moments when the body passes point at height . Find the initial speed and total motion time .

Answer

; .

Hint

See problem 38. Solve the problem using another method.

46. From points and , vertically aligned ( above) and apart, two bodies are thrown simultaneously at : from downward, from upward. When and where will they meet?

Answer

5 s; 75 m below point B.

Hint

The relative speed of the bodies is constant and equal to . Both bodies approach each other with this speed. Therefore, the time to meeting is s.

47. A body is thrown upward with initial speed . At its peak, a second body is thrown from the same point at the same speed. At what height will they meet?

Answer

Hint

The problem can be solved beautifully if you realize that the moment of meeting divides the ascent time, as well as the fall time of the bodies, into equal parts. The paths traveled by a body falling without initial velocity in successive equal time intervals relate as 1:3. Therefore, the meeting will occur at 3/4 of the height.

48. Two bodies are thrown upward from the same point at , with a time interval . After what time and at what height will they meet?

Answer

s; m.

Hint

See problem 38. Solve the problem graphically as well.

49. A balloon rises vertically from Earth at . After , an object falls out. How long until it hits the ground?

Answer

Solution

Let's take the coordinate axis directed vertically upwards, with the origin on the Earth's surface. Then the kinematic equation of motion of the object has the form where and Substituting the coordinate value of the object at the moment of fall into this equation, we find .

50. A balloon descending at speed throws an object upward at speed relative to Earth. What will be the distance between them when the object reaches its peak? What is the maximum distance ? How long until they are level again?

Answer

; ; .

Hint

The kinematic equation of motion of the body is conveniently written in the coordinate system associated with the moving aerostat. Here is the initial velocity of the body; is the acceleration; is the coordinate of the body when the velocity of the body relative to the aerostat is zero; is its coordinate at the moment , when its velocity relative to the Earth is zero. Since , then at . At , the coordinate of the body is zero.

51. A body at point , above Earth, starts falling freely. Simultaneously, another body is thrown upward from point , below , at speed . Find if both bodies hit the ground simultaneously. Neglect air resistance. Take .

Answer

m/s.

Solution

1st method. Taking the Earth's surface as the origin, let's write the equations of motion for both bodies:

At the moment of landing . Substituting the values and into equations (1) and (2) and solving them for , we find

2nd method. Both bodies approach each other, moving uniformly relative to each other. The initial distance between them is . Therefore, the velocity can be found from the equation of uniform motion , where is the fall time of the first body from height H. Since . We find .

52. A body falls freely from height . Simultaneously, another body is thrown downward from height . Both hit the ground simultaneously. Find the initial speed of the second body. Verify with , , and .

Answer

m/s.

53. A stone is thrown horizontally from a mountain slope at angle . What speed is needed for the stone to land on the mountain at distance from the peak?

Answer

Solution

Writing the coordinate equations for the stone (see figure): and considering that at the moment of fall of the stone we get after eliminating , that \vec{v}_{\text{rel}} = \vec{u} - \vec{v} = \vec{u}_0 - \vec{v}_0.v_{0x} = v_0 \cos \alpha,\quad v_{0y} = v_0 \sin \alpha.t = l/v_{0x} = l/(v_0 \cos \alpha).\tag{1}H-h = gt^2/2, \tag{2}h = v_{0y} t - gt^2/2 = v_0 \sin \alpha t - gt^2/2. \tag{3}H = t v_0 \sin \alpha. \tag{4}\tan \alpha = \dfrac{H}{l},\tan \alpha = {l}. \tag{4}t = \sqrt{\dfrac{2l}{a}} = \sqrt{\dfrac{2d}{g}}.H_{\text{max}} =\dfrac{v_{0y}^2}{2g}, \tag{1}v_0^2 = v_x^2 + v_{0y}^2, \tag{2}v^2 = v_x^2 + (v_{0y} - gt)^2, \tag{3}v_{0y} = \dfrac{v_0^2 - v^2 + g^2 t^2}{2gt}.\tag{4}H_{\text{max}} = \dfrac{(v_0^2 - v^2 + g^2 t^2)^2}{8g^3 t^2}.x = v_0 t \cos \beta,\quad y = v_0 t \sin \beta - gt^2/2.v_0 = \sqrt{\dfrac{gL \cos \alpha}{2 \cos \beta \sin(\beta - \alpha)}}.v_0 = \sqrt{\dfrac{gl \cos^2 \alpha}{2 \cos \beta \sin(\beta - \alpha)}}.l = \dfrac{2 v_0^2 \cos \beta \sin(\beta - \alpha)}{g \cos^2 \alpha}.\cos \beta_0 \sin(\beta_0 - \alpha) = \dfrac{1}{2} [\sin(2\beta_0 - \alpha) - \sin \alpha].\sin(2\beta_0 - \alpha) = 1\beta_0 = \dfrac{\pi}{4} + \dfrac{\alpha}{2}y = v_{0y} t + a_y t^2 / 2,v_{0y} = v_0 \cos \alpha\text{ and } a_y = -g \cos \alphat = 2v_{0y}/a_y = 2v_0/g = 2\sqrt{2gh}/g. \begin{aligned} s_1 &= v_0 \dfrac{2v_0}{g} \sin \alpha + \dfrac{g}{2} (\dfrac{2v_0}{g})^2 \sin \alpha \\ &= \dfrac{4v_0^2}{g} \sin \alpha \\ &= 8H \sin \alpha; \end{aligned} v_{2y} = v_{0y} - a_y t = v_0 \cos \alpha - g \dfrac{2v_0}{g} \cos \alpha = -v_{0y}; \begin{aligned} v_{2x} &= v_{0x} + a_x t \\ &=v_0 \sin \alpha + g \dfrac{2v_0}{g} \sin \alpha \\ &= 3v_0 \sin \alpha, \end{aligned} v_{ny} = -v_{0y},v_{nx} = (2n-1)v_0 \sin \alpha.2v_0 \sin \alpha,\quad 4v_0 \sin \alpha,\quad 6v_0 \sin \alpha,s_1:s_2:s_3... = 1:2:3.\tau = \dfrac{h}{c} + \sqrt{\dfrac{2h}{g}}.\tag{1}\dfrac{\Delta h}{h} =\dfrac{h_1 - h}{h} = 0.05.h = 0.95 h_1 = 0.95 \dfrac{g\tau^2}{2}.\tau = \dfrac{2(1 - 1/\sqrt{0.95})c}{0.95 g} \approx 1.77\text{ s}.$

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