Movement in a straight line with constant acceleration examples of problem solving
From paragraphs A And B, the distance between which is l, two bodies began to move towards each other simultaneously: the first with a speed v 1 , second - v 2. Determine how long they will meet and the distance from the point A to their meeting point. Solve the problem graphically.
Solution
1st way:
The dependence of the coordinates of bodies on time:
At the moment of meeting, the coordinates of the bodies will coincide, i.e. . This means that the meeting will take place after the time from the beginning of the movement of the bodies. Find the distance from the point A to the meeting point as .
2nd way:
The velocities of the bodies are equal to the tangent of the slope of the corresponding graph of the dependence of the coordinate on time, i.e.,. The moment of the meeting corresponds to the point C graph intersections.
After what time and where would the bodies meet (see problem 1), if they moved in the same direction A→B, and from the point B the body began to move through t 0 seconds after the start of its movement from the point A?
Solution
Graphs of the dependence of the coordinates of bodies on time are shown in the figure.
Based on the figure, we will compose a system of equations:
Having solved the system with respect to 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 down the river in time t 1 = 3 hours, and the raft is in time t= 12 hours What time t 2 will the motorboat cost for the return journey?
Solution
Let be s- distance between points A And B, v is the speed of the boat in relation to the water, and u- flow rate. Expressing the distance s three times - for a raft, for a boat moving with the current, and for a boat moving against the current, we get a system of equations:
Solving the system, we get:
The subway escalator lowers a person walking down it in 1 minute. If a person walks twice as fast, he will descend in 45 seconds. How long does the person standing on the escalator go down?
Solution
Denote by letter l escalator length; t 1 is the descent time of a person walking at a speed v; t 2 is the descent time of a person walking at a speed of 2 v; t- the time of the 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, with speed 2 v and stands motionless on the escalator), we get a system of equations:
Solving this system of equations, we get:
A man runs up the escalator. The first time he counted n 1 \u003d 50 steps, the second time, moving in the same direction at a speed three times greater, he counted n 2 = 75 steps. How many steps would he count on a stationary escalator?
Solution
Since, with an increase in speed, a person counted a greater number of steps, it means that the directions of the velocities of the escalator and the person coincide. Let be v is the person's speed relative to the escalator, u- escalator speed, l- the length of the escalator, n is the number of steps on a fixed escalator. The number of steps that fit in a unit length of the escalator is n/l. Then the time spent by a person on the escalator when he moves relative to the escalator at a speed v equals l/(v+u), and the path taken along the escalator is equal to vl/(v+u). Then the number of steps on this path is equal to . Similarly, for the case when the speed of a person relative to the escalator is 3 v, we get .
Thus, we can compose a system of equations:
Eliminating the ratio u/v, we get:
Between two points located on the river at a distance s\u003d 100 km from one another, a boat runs, which, going downstream, covers this distance in time t 1 \u003d 4 hours, and against the current, - for the time t 2 = 10 hours Determine the speed of the river u and boat speed v regarding water.
Solution
Expressing the distance s twice, for a boat going downstream and 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 the village, located at a distance s 1 = 15 km from the pier, a motorboat leaves down the river. She reached the village in time t= 3/4 h and, turning back, met the raft at a distance s 2 = 9 km from the village. What is the speed of the river and the speed of the boat through the water?
Solution
Let be v- boat speed u is the speed of the river. Since from the moment of departure of the motor boat from the pier to the moment of the meeting of the motor boat with the raft, obviously, the same time will 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 before the meeting, for a raft, and on the right is for a motor boat. Let's write an equation for the time that the motorboat spent to overcome the path s 1 from the pier to the village: t=s 1 /(v+u). Thus, we obtain a system of equations:
Where do we get v= 16 km/h, u= 4 km/h.
A column of troops during a campaign moves at a speed v 1 = 5 km / h, stretching along the road for a distance l\u003d 400 m. The commander, who is at the tail of the column, sends a cyclist with an order to the head detachment. The cyclist sets off and rides at a speed v 2 \u003d 25 km / h and, having completed the order on the go, immediately returns back at the same speed. After how much time t after receiving the order, he returned back?
Solution
In the frame of reference associated with the column, the speed of the cyclist when moving towards the vanguard is v 2 -v 1 , and when moving back v 2 +v one . That's why:
Simplifying and substituting numerical values, we get:
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Wagon 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 car walls relative to each other is equal to l\u003d 6 cm. What is the speed of the bullet?
Solution
Denote by letter u bullet speed. The time of flight of a bullet from the wall to the wall of the car is equal to the time for which the car covers the distance l. Thus, we can write an equation:
From here we find u:
.
What is the speed of the drops v 2 sheer falling rain, if the driver of a car noticed that the raindrops do not leave a trace on the rear window tilted forward at an angle α = 60° to the horizon when the vehicle speed v 1 over 30 km/h?
Solution
As can be seen from the figure,
so that raindrops do not leave a trace 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 it takes the car to cover the distance l:
Or, expressing from here v 2:
It is raining outside. In which case will a bucket standing in the back of a truck fill faster with water: when the car is moving or when it is standing still?
Answer
Equally.
At what speed v and at what course should the plane fly so that in time t= 2 hours to fly exactly to the North path s= 300 km if during the flight a northwest wind blows at an angle α = 30° to the meridian with speed u= 27 km/h?
Solution
We write down the system of equations according to the figure.
Since the plane must fly due north, the projection of its speed on the axis Oy v y is y- wind speed component u y .
Having solved this system, we find that the aircraft should keep its course to the northwest at an angle of 4 ° 27 "to the meridian, and its speed should be equal to 174 km / h.
Moves along a smooth horizontal table with a speed v Black board. What shape will the chalk leave on this board if it is thrown horizontally with a speed of u perpendicular to the direction of movement of the board, if: a) the friction between the chalk and the board is negligible; b) is there a lot of friction?
Solution
The chalk will leave a mark on the board, which is a straight line that makes an angle arctg ( u/v) with the direction of movement of the board, i.e., coincides with the direction of the sum of the velocity 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 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, constituting the angle α with line AB.
At what angle β to the line AB should have been omitted from paragraph B torpedo to hit the ship? The torpedo must be launched at the moment when the ship was at the point A. The speed of the torpedo is u.
Solution
Dot C in the figure - this is the meeting point of the ship and the torpedo.
AC = vt, BC = ut, where t- time from start to meeting. According to the sine theorem
From here we find β :
.
To the slider, which can move along the guide rail,
a cord is attached, 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.
For a very short period of time Δt the slider moves a distance AB = Δl.
The cord for the same period of time is chosen for the length AC = 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 pulling out the rope is equal to the projection of the speed of the crawler on the direction of the rope.
Workers lifting a load
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.
Load speed projection u per 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 rails.
Coupling A moving at a constant speed v A = 30 cm/s. Find speed v B clutch B when the angle OAB= 60°. Taking as the beginning of the time reference the moment when the clutch A was at the point O, determine the distance OB and clutch speed B in a function of time.
Answer and solution
v B= v A ctg α
= 17.3 cm/s; , 
.
At any point in time, the velocity projections v A and v B rod ends
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 = v B sin α . Where v B = v A ctg α .
At any point in time for a triangle OAB the Pythagorean theorem is valid: l 2 = OA 2 (t) + OB 2 (t). Let's find it from here OB(t): . Insofar as OA(t) = v A t, then we finally write the expression for OB(t) So: .
Because ctg α
at any moment is equal to OA(t)/OB(t), then we can write the expression for the dependence v B from time: .
The tank is moving at a speed of 72 km/h. With what speed do they move relative to the Earth: a) the upper part of the caterpillar; b) the lower part of the caterpillar; c) the point of the track that is currently moving vertically with respect to the tank?
Answer and solution
a) 40 m/s; b) 0 m/s; c) ≈28.2 m/s.
Let be v- the speed of the tank relative to the Earth. Then the speed of any point of the caterpillar relative to the tank is also equal to v. The speed of any point of the caterpillar relative to the Earth is the sum of the vectors of the tank's velocity relative to the Earth and the velocity of the caterpillar's point 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 way at a speed v 1 = 40 km / h, the second - at a speed v 2 = 60 km/h. Find the average speed for the entire distance travelled.
2. The car traveled half way at a speed v 1 \u003d 60 km / h, the rest of the way he walked half the time at a speed v 2 \u003d 15 km / h, and the last section - with a speed v 3 = 45 km/h. Find the average speed of the car for the entire journey.
Answer and solution
1. v cf =48 km/h; 2. v cf = 40 km/h.
1. Let s- all the way t- the time spent on overcoming the entire path. Then the average speed for the entire journey is s/t. Time t consists of the sum of the time intervals spent on overcoming the 1st and 2nd halves of the path:
Substituting this time into the expression for the average speed, we get:
.(1)
2. The solution of this problem can be reduced to the solution (1.), if we first determine the average speed on the second half of the journey. Let's call this speed v cp2, then we can write:
where t 2 - the time spent on overcoming the 2nd half of the journey. The path traveled during this time consists of the path traveled at a speed v 2 , and the path traveled at a speed v 3:
Substituting this into the expression for v cp2 , we get:
.
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The train traveled for the first half of the journey at a speed of n\u003d 1.5 times greater than the second half of the path. The average speed of the train for the whole journey v cp = 43.2 km/h. What are the speeds of the train on the first ( v 1) and second ( v 2) half way?
Answer and solution
v 1 =54 km/h, v 2 =36 km/h.
Let be t 1 and t 2 - time for the train to pass the first and second half of the journey, respectively, s- the entire distance traveled by the train.
Let's make 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 =n.v. 2 and solving the resulting system of equations, we obtain v 2 .
Two balls began to move simultaneously and with the same speed on 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 time of movement of the first ball will be longer.
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 path traveled), therefore, as can be seen from the figure, t 1 >t 2 .
The plane flies from the point A to paragraph B and returns to point A. The speed of the aircraft 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
Aircraft flight time from point A to paragraph B and back when the wind blows along the line AB:
.
Then the average speed in this case:
.
If the wind blows perpendicular to the line AB, the aircraft velocity vector must be directed at an angle to the line AB so as to compensate for the influence of the wind:
The round-trip flight time in this case will be:
Aircraft flight speed per point B and vice versa are identical and equal:
.
Now we can find the ratio of the average velocities obtained for the considered cases:
.
Distance between two stations s= 3 km the metro train passes at an average speed v cf = 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. Draw a graph of train speed and determine the highest speed of the train v Max.
Answer and solution
The figure shows a graph of the speed of the train.
The distance traveled by the train is numerically equal to the area of the figure bounded by the graph 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:
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The last car is unhooked from the moving train. The train continues to move at the same speed v 0 . How will the paths covered by the train and the car relate to the moment the car stops? Assume that the car was moving with uniform speed. Solve the problem graphically.
Answer
At the moment when the train started, the person seeing off began to run uniformly along the course of the train with a speed v 0 =3.5 m/s. Assuming the movement of the train is uniformly accelerated, determine the speed of the train v at the moment when the escort catches up with the escort.
Answer
v=7 m/s.
A graph of the dependence of the speed of some body on time is shown in the figure.
Draw graphs of the dependence of the acceleration and coordinates of the body, as well as the distance traveled by it from time.
Answer
Graphs of the dependence of acceleration, the coordinates of the body, as well as the distance traveled by it from time are shown in the figure.
The graph of the dependence of the acceleration of the body on time has the form shown in the figure.
Draw graphs of the speed, displacement and distance traveled by the body versus time. The initial velocity of the body is equal to zero (acceleration is equal to zero in the section of the discontinuity).
The body starts moving from a point A with speed v 0 and after some time hits the point B.
What distance did the body go 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 dependence of the coordinate of the body on time.
after a moment t=t 1 graph curve - parabola. What is the movement shown in this graph? Construct a graph of the speed of the body as a function of 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: equally 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 body's velocity versus time.
The figure shows velocity graphs for two points moving along the same straight line from the same initial position.
Known time points t 1 and t 2. At what point in time t 3 dots meet? Build motion graphs.
In what second from the beginning of the motion, the path traveled by the body in uniformly accelerated motion is three times the distance traveled in the previous second, if the motion occurs without an initial velocity?
Answer and solution
For the second second.
The easiest way to solve this problem graphically. Because the path traveled by the body is numerically equal to the area of the figure under the line of the velocity graph, then it is obvious from the figure 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 goods in the shortest possible time from one place to another, located 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 load in the minimum time if it moves with acceleration for the first half of the way + a, and the remaining half with acceleration - a.
Then the following expressions can be written: L = ½· vt 1 ; v = ½· at 1 ,
where we find the maximum speed:
A jet plane is flying at a speed v 0 =720 km/h. From a certain moment the plane moves with acceleration for t\u003d 10 s and at the last second the path passes s\u003d 295 m. Determine the acceleration a and final speed v aircraft.
Answer and solution
a\u003d 10 m / s 2, v=300 m/s.
Let's plot the speed of the aircraft in the figure.
Aircraft speed at time t 1 equals v 1 = v 0 + a(t 1 - t 0). Then the path traveled by the aircraft in the time from t 1 to t 2 equals s = v 1 (t 2 - t 1) + a(t 2 - t 1)/2. From this we can express the desired value of acceleration a and, substituting the values from the condition of the problem ( 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\u003d 10 m / s 2. final speed of the aircraft 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 t 1 \u003d 1 s, and the second - for t 2 = 1.5 s. Wagon length l=12 m. Find the acceleration a trains and their speed v 0 at the start of the observation. The movement of the train is assumed to be equally variable.
Answer and solution
a\u003d 3.2 m / s 2, v 0 ≈13.6 m/s.
The distance traveled by the train so far t 1 is:
and the path to the point in time t 1 + t 2:
.
From the first equation we find v 0:
.
Substituting the resulting expression into the second equation, we obtain the acceleration a:
.
A ball thrown up an inclined plane passes successively two equal segments of length l each one keeps moving on. The first segment of the ball went for t seconds, the second - for 3 t seconds. Find speed v ball at the end of the first segment of the path.
Answer and solution
Since the considered movement of the ball is reversible, it is advisable to choose the common point of the two segments as the starting point. In this case, the acceleration during movement on the first segment will be positive, and when moving on the second segment, it will be negative. The initial speed in both cases is equal to v. Now let's write down the system of equations of motion for the paths traveled by the ball:
Eliminating acceleration a, we get the desired speed v:
A board divided into five equal segments begins to slide down an inclined plane. The first segment went past the mark made on the inclined plane in the place where the leading edge of the board was at the beginning of the movement, beyond τ =2 s. How long will it take for the last segment of the board to pass this mark? The motion of the board is assumed to be uniformly accelerated.
Answer and solution
τ n = 0.48 s.
Find the length of the first segment:
Now we write down the equations of motion for the points of origin (time t 1) and end (time 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 flying at a speed of 400 m / s hits an earthen rampart and penetrates it to a depth of 36 cm. How long did it move inside the rampart? With what acceleration? What was its speed at a depth of 18 cm? At what depth did the speed of the bullet decrease three times? The movement is assumed to be uniform. 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.
The time of movement of a bullet inside the shaft is found 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 is rolled up an inclined board. On distance l= 30 cm from the beginning of the path, the ball visited twice: through t 1 = 1 s and after t 2 = 2 s after the start of movement. Determine initial speed v 0 and acceleration a the motion of the ball, assuming it to be constant.
Answer and solution
v 0 = 0.45 m/s; a\u003d 0.3 m / s 2.
The dependence of the ball speed on time is expressed by the formula v = v 0 - at. At the point in time t = t 1 and t = t 2 the ball had the same magnitude and opposite speeds: v 1 = - v 2. But v 1 =v 0 - at 1 and v 2 = v 0 - at 2 , so
v 0 - at 1 = - v 0 + at 2 , or 2 v 0 = a(t 1 + t 2).
Because the ball is moving with uniform acceleration, the distance l can be expressed as follows:
Now you can make a system of two equations:
,
solving which we get:
A body falls from a height of 100 m with no initial velocity. How long does it take the body to cover the first and last meters of its path? What path does the body cover in the first, in the 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 of the open position of the photographic shutter τ , if when photographing a ball falling along the vertical centimeter scale from the zero mark without 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 0.5 s. Find the height of the fall.
Answer
A freely falling body has traveled 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 down the ball from a height h so that he jumps to height 2 h? Neglect air friction and other mechanical energy losses.
Answer
With what time interval did two drops break away from the roof eaves, if two seconds after the second drop began to fall, the distance between the drops was 25 m? Ignore air friction.
Answer
τ ≈ 1 s.
The body is thrown vertically upwards. The observer notices the time t 0 between two times when the body passes the point B at the height h. Find the initial throwing speed v 0 and the time of the whole body movement t.
Answer
; .
From points A And B located vertically (point A above) at a distance l\u003d 100 m apart, two bodies are thrown simultaneously with the same speed of 10 m / s: from A- vertically down B- vertically up. When and where will they meet?
Answer
t= 5 s; 75 m below the point B.
A body is thrown vertically upwards with an initial velocity v 0 . When it reached the highest point of the path, from the same starting point with the same speed v 0 the second body is thrown. At what height h from the starting point will they meet?
Answer
Two bodies are thrown vertically upward from the same point with the same initial velocity 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 bodies meet?
Answer
t= 1.75 s; h≈ 19.3 m.
The balloon rises from the Earth vertically upwards with acceleration a\u003d 2 m / s 2. Across τ = 5 s from the beginning of its movement, an object fell out of it. After how much time t will this object fall to the ground?
Answer
t≈ 3.4 s.
From a balloon descending at a speed u, throw up the body with a speed v 0 relative to the Earth. What will be the distance l between the balloon and the body by the time of the highest rise of the body relative to the Earth? What is the longest distance l max between body and balloon? After what time τ from the moment of throwing the body catches up with the balloon?
Answer
l = v 0 2 + 2UV 0 /(2g);
l max = ( u + v 0) 2 /(2g);
τ = 2(v 0 + u)/g.
body 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 upwards. 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. Ignore air resistance. Accept g\u003d 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 hit the ground at the same time. Determine initial speed v 0 of the second body. Check the correctness of the solution on a numerical example: h= 10 m, H= 20 m Accept g\u003d 10 m / s 2.
Answer
v 0 ≈ 7 m/s.
A stone is thrown horizontally from the top of a mountain with slope α. At what speed v 0 a stone must be thrown for it to fall on a mountain in the distance L from the top?
Answer
Two people play ball by throwing it to each other. What is the maximum height the ball reaches during the game if it flies from one player to another for 2 s?
Answer
h= 4.9 m.
The aircraft is flying at a constant altitude h in a straight line at a speed v. The pilot must drop the bomb at 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 aircraft is located at this moment? The air resistance to the movement of the bomb is ignored.
Answer
; .
Two bodies fall from the same height. On the path of one body there is an area located at an angle of 45 ° to the horizon, from which this body is elastically reflected. How do the times and velocities of the fall of these bodies differ?
Answer
The time of the fall of the body, on the path of which the platform was located, is longer, since the vector of the velocity gained by the moment of the collision changed its direction to the horizontal one (during an elastic collision, the direction of the velocity changes, but not its magnitude), which means that the vertical component of the velocity vector became equal to zero, while as for another body, the velocity vector 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 that 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 fell 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 ∆h, 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 bodies move freely near the Earth's surface, their relative speed is constant.
From a point A 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 angle α does not depend on initial speed v 0 body thrown from a point B, and determine this angle if . Ignore air resistance.
Answer
α = 60°.
Body thrown at an angle α to the horizon at a speed v 0 . Determine speed v this body is on top h over the horizon. Does this speed depend on the angle of throw? Air resistance is ignored.
at an angle α =60° to the horizon a body is thrown with an initial velocity v=20 m/s. After how much time t it will move at an angle β =45° to the horizon? There is no friction.
From three pipes located on the ground, water jets hit at the same speed: at an angle of 60, 45 and 30 ° to the horizon. Find ratios of greatest heights h the rise of the water jets flowing from each pipe and the fall distances l water to the ground. Air resistance to the movement of water jets is not taken into account.
From a point lying at the upper end of the vertical diameter d of some circle, along the grooves installed along the various chords of this circle, loads simultaneously begin to slide without friction.
Determine how much time t weights reach the circumference. How does this time depend on the angle of inclination of the chord to the vertical?
Initial speed of the thrown stone v 0 =10 m/s, and later t\u003d 0.5 s stone speed v=7 m/s. To what maximum height above the initial level will the stone rise?
Answer
H max ≈ 2.8 m.
At a certain height, balls are ejected simultaneously from one point with the same speed in all possible directions. What will be the locus of the balls at any given time? Ignore air resistance.
Answer
The geometric location of the points of location of the balls at any time will be a sphere, the radius of which v 0 t, and its center is located below the starting point by an amount gt 2 /2.
A target located on a hill is visible from the location of the gun at an angle α to the horizon. 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. Air resistance is ignored. At what elevation angle β 0 firing range along the slope will be the 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 target coordinates ( x = L, y = L tgα) and eliminating t, we get:
Range l projectile flight along the slope l = L/ cos α . Therefore, the formula that we received can be rewritten as follows:
,
this expression is maximum at the maximum value of the product
That's why l maximum at maximum value = 1 or
At α = 0 we get a response β 0 = π /4 = 45°.
An elastic body falls from a height h on an inclined plane. Determine how long t After reflection, the body will fall on an inclined plane. How does time depend on the angle of the inclined plane?
Answer
It does not depend on the angle of the inclined plane.
From high H on an inclined plane that forms an angle with the horizon α \u003d 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 terms (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 could be the error τ in determining the moments of the shot and the arrival of the 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 noticing the time τ through which the splash will be heard. Starting from what values τ is it necessary to take into account the transit time of the sound? speed of sound in air c=330 m/s.
Answer
Most problems for the motion of bodies with constant acceleration are solved basically in the same way as problems for uniform rectilinear motion (see § 1.9). However, instead of one equation for the dependence of the coordinate on time, there will now be two: for the coordinate and for the projection of the velocity depending on time:
2 "
X = Xq + v0xt +
2? Task 1
The skater, having accelerated to a speed of v0 = 6 m/s, began to glide uniformly. After a time t = 30 s, the speed modulus of a skater moving in a straight line became equal to v = 3 m/s. Find the skater's acceleration, assuming it to be constant.
Solution. Let's align the X-axis with the trajectory of the skater. For the positive direction of the axis, we choose the direction of the initial velocity vector v0 (Fig. 1.66). Since the skater moves with
constant acceleration, then vx = v0x + axt. Hence ah = , where
vx = v and vQx = v0, since the vectors 50 and v have the same direction
v-v0
lower than the X axis. Therefore, ax = ---, ax = -0.1 m/s2 and
a = 0.1 m/s2. The minus sign indicates that the acceleration is opposite to the X axis.
Task 2
A bar on a smooth inclined plane was given an initial velocity v0 = 0.4 m/s directed upwards. The bar moves in a straight line with constant acceleration, the module of which is a = 0.2 m/s2. Find the velocities of the bar at times equal to 1, 2, 3 s from the start of motion. Determine the position of the bar at these moments of time relative to the point where the bar had a speed u0. What is the distance traveled by the block in 3 seconds?
Solution. The acceleration of the bar is directed downward along the plane both during its ascent and during its descent.
97
4-Myakishev, 10 cells
Let's combine the coordinate axis with the trajectory of motion. For the positive direction of the X axis, we take the direction of the initial velocity vector u0. We choose the origin of coordinates at that point of the trajectory where the bar had a speed v0 (Fig. 1.67).? The block moves with constant acceleration, so vx = vQx + axt. Since v0x = vQ, ax = -a, then their = v0 - at. This formula is valid for any moment in time.
Let's find the projections and modules of velocities at the indicated moments of time:
vlx = v0 - atl = 0.2 m/s, vx = |uljt| = 0.2 m/s;
v2x = v0- at2 = 0, v2 = 0;
v3x = v0 - at3 = -0.2 m/s, v3 = |u3J = 0.2 m/s.
Since vlx > 0, the velocity is directed in the same direction as the X axis. The minus sign of the v3x projection indicates that the velocity v3 is directed in the direction opposite to the X axis. This is how it should be, because after stopping ( v2 = 0) the block will start to slide down the plane.
Let's find the position of the bar for the given moments of time:
.2
at\ _ . 0.2 m _ 0 x1 \u003d v0t1 - \u003d 0.4 m - - \u003d 0.3 m,
.2at2
x2 \u003d v0t2 - -g- \u003d 0.8 m - 0.4 m \u003d 0.4 m,
.2at3
x3 \u003d v0t3 - -g- \u003d 1.2 m - 0.9 m \u003d 0.3 m.
Pay attention to the fact that at point B with a coordinate of 0.3 m (lg1 \u003d lg3) (see Fig. 1.67) the body was twice (during ascent and descent). At the same time moments, the body had velocities equal in absolute value (L>1 = L>3), but opposite in direction: v1 - -v3.
At point A with coordinate x2 (see Fig. 1.67) the speed is v2 = 0. Here there was a change in the direction of the speed. At time t3 = 3 s, the bar was at point B with coordinate x3. Therefore, the path traveled by the bar
s - OA + AB \u003d 2X2 - x3 \u003d 0.5 m.
Task 3
Figure 1.68, a shows a graph of the projection of the speed of a point on time. Plot the coordinate versus time graph, if the initial coordinate i, = 5 m, Plot the path versus time plot.
Solution. First, let's build a graph of coordinates versus time. For the first 2 s, the point moved uniformly in the opposite direction to the X axis (vlx B the next 2 s, the movement was uniformly accelerated in the same direction as at the beginning (v2x
Yu t, s
From 4 to 6 s, the point again moved uniformly in the same direction, so x3 = x2 + Lx3 = -1 m - 3 m = -4 m. The graph is a parabola Сwhere Dl is its top.
8 Yu t, s
From 6 to 8 s, the point moved with uniform acceleration in the positive direction of the X axis (v4x > 0). Graph - parabola DXEj. By the end of the 8th second, the coordinate Ї4 = -4M + ZM = -1 M. Further, the point moved uniformly in the same direction (v5x > 0): = -1 m + 3 m = 2 m. Graph - parabola E1FV? 1. When constructing a path graph, it must be taken into account that the path is a non-negative value and cannot decrease in
the process of movement.
The graph consists of segments of parabolas A2B2, B2C2, C2D2, D2E2, E2F2 (Fig. 1.68, c).
Exercise 3
A small cube on a smooth inclined plane was given an initial velocity u0 = 8 m/s directed upwards. The cube moves in a straight line with constant acceleration, the module of which is a = 2 m/s2. Find the position of the cube relative to the point of the plane where the velocity v0 is given to the cube, at the moments of time 2, 4, 6 s from the beginning of the movement, as well as the speed of the cube at the same moments of time. What is the distance traveled by the cube in 5 seconds?
Two cyclists are riding towards each other. One of them with an initial speed of 18 km/h rises uphill uniformly with constant acceleration, the module of which is 20 cm/s2. Another cyclist with an initial speed of 5.4 km/h descends the mountain with the same acceleration modulo. How soon will they meet? At what distance from the foot of the mountain will the meeting take place and what path will each of them have traveled by this moment? The distance between the cyclists at the initial moment of time was 195 m.
Figure 1.69 shows graphs I, II and III of projections of the velocity of three bodies moving in a straight line. Describe the features of the movement of bodies. What corresponds to point A of the intersection of graphs? Find the acceleration modules of bodies. Write down the formulas for calculating the projections of the speed of each body.
A train travels a distance of 20 km between two stations at a speed whose average modulus is 72 km/h, and it spends 2 minutes on acceleration, and then goes at a constant speed. The train takes 3 minutes to decelerate to a complete stop. Determine the modulus of the train's maximum speed.
The sled rolling down the mountain travels 2 m in the first 3 s, and 4 m in the next 3 s. Considering the movement to be uniformly accelerated, find the module of acceleration and the module of the initial speed of the sled.
A body moving uniformly accelerated with an initial speed of 1 m/s acquires, after passing some distance, a speed of 7 m/s. What was the speed of the body at the middle of this distance? Vx, m/s
vx> m/s
-4"
Rice. 1.70
4
ABOUT
Rice. 1.69
t, s A point begins to move along a straight line with constant acceleration. After a time t1 after the start of its movement, the direction of the point's acceleration changes to the opposite, remaining unchanged in absolute value. Determine how much time t2 after the start of movement
The point will return to its original position.
The trolley must transport the load in the shortest possible time from one place to another, remote from the first by a distance L. It can increase or decrease its speed only with the same acceleration equal to a. In addition, it can move at a constant speed. What is the maximum speed modulo the trolley must reach in order to fulfill the above condition?
Figure 1.70 shows a graph of the projection of the speed of a point moving in a straight line, as a function of time. Plot the coordinate versus time if = 4.5 m. Plot the path versus time.
1. The body moves with constant acceleration and zero initial speed. Show graphically that the paths traveled by the body in successive equal time intervals are related as successive odd numbers.
Solution . With a uniformly accelerated motion of a body with zero initial speed, its speed over time t changes by law
where a- acceleration.
Let's build a speed graph (see Fig.) and mark on the axis t equal intervals OA 1 =BUT 1 BUT 2 =BUT 2 BUT 3 =BUT 3 BUT 4 = ...; from points BUT 1 ,BUT 2 , … draw vertical lines with a dotted line until they intersect with the speed graph at points IN 1 ,IN 2 ,IN 3 , … . Then the path traveled during the first interval is numerically equal to the area of the triangle OA 1 IN one ; the paths traveled over subsequent intervals are equal to the areas of the corresponding trapezoids. The graph shows that the area of the first trapezoid BUT 1 BUT 2 IN 2 IN 1 is three areas of a triangle OA 1 IN one ; area of the next trapezoid BUT 2 BUT 3 IN 3 IN 2 equals five areas of a triangle OA 1 IN 1 etc. Therefore, the ratio of the paths traversed by the body in successive equal time intervals is equal to:
S 1:S 2:S 3: …: S n = 1:3:5: …: (2n – 1).
2. In the fifth second of uniformly accelerated motion with zero initial velocity, the body travels a path S 2 = 36 m. Which way S 1 passes the body in the first second of this movement?
Solution . From the solution of the previous problem, it follows that
S 1:S 5 = 1:9.
Consequently,
4 m
3. A freely falling body has traveled 1/3 of its path in the last second of its fall. Find the fall time t and height h from which the body fell.
Solution . From the laws of motion of a body with constant acceleration and zero initial velocity, we obtain the following equations:
Here = 1 s. Solving the resulting system of equations, we find:
According to the task t> 1. This condition is satisfied by the root 
5.4 s Next we get:
4. The balloon rises vertically upwards from the surface of the Earth with acceleration a = 2 m/s 2 . After = 10 s after the start of movement, an object came off the basket of the ball. What is the maximum height h m will this item go up? After what time t 1 and with what speed v 1 will it fall to the Earth?
R 
solution
.
The object came off the basket of the balloon at a height 
having speed v 0 = but pointing vertically upwards. Let's choose a reference system - the axis OH, directed vertically upwards, and depict in the figure the position of the object at the moment of separation from the basket. The maximum height is
h m =h 0 +S m ,
where 
- the path traveled by the object during the time after the liftoff until the rise to the maximum height, i.e.
Further, it is obvious that after the separation, the object moves upward during the time 
until it stops at its highest point, after which it falls freely from a height h m; while the time of its fall tfind from the relation 
those. 
Consequently,
The speed of an object that fell to the Earth is determined from the relation
5. With what time interval did two drops of water break away from the roof eaves if, two seconds after the second drop began to fall, the distance between them was S= 25 m?
Solution . Let be the time interval between the separation of the first and second drops, t= 2 s - time from the moment of separation of the second drop. Then, by the time the second drop breaks off, the first drop has passed the path S 0 = g 2 /2 and had a speed v 0 = g. Further, it is obvious that the distance between the drops is equal to
where 
- the path traveled by the first drop in time t,
- the path traveled by the second drop during the same time.
Consequently,
Solving the resulting equation and taking into account that > 0, we find:
6. A ball is rolled up an inclined board. On distance l\u003d 30 cm from the beginning of the throw, the ball visited twice: after t 1 = 1 s and after t 2 = 2 s after the start of movement. Determine initial speed v 0 and acceleration a ball, assuming it to be constant.
Solution . We write down the law of motion of the ball, choosing the axis OX, directed along the movement of the ball:
Let's rewrite this equation like this:
At x=l this equation has roots t 1 and t 2 .
Therefore, by Viette's theorem
Solving this system, we find:
\u003d 30 cm / s 2,
= 45 cm/s.
Comment
. This problem can be solved in another way, namely: using the law of motion 
write two equations x(t 1) =l And x(t 2) =l, and then solve the resulting system of equations with two unknowns v 0 and a.