On the straight starts the point with constant. Movement in a straight line with constant acceleration Examples of solving problems
From points A. and B., the distance between which is equal l., at the same time, two bodies began to move each other: the first at speed v. 1, second - v. 2. Determine how much time they will meet and distance from the point A. to the place of their meeting. Solve the task also graphically.
Decision
1st way:
Dependence of coordinates of bodies from time to time:
At the time of the meeting, the coordinates of the bodies coincide, that is,. It means that the meeting will occur in time from the beginning of the body. We will find the distance from the point A. Before the meeting place as.
2nd way:
Tel velocities are equal to tangent angle of inclination of the corresponding graph of the dependence of the coordinate from time, i.e.,. The moment of the meeting corresponds to the point C. Crossing graphs.
What time and where they would meet the body (see Task 1), if they moved in the same direction A.→B., and from the point B. body start to move through t. 0 seconds after the start of movement from the point A.?
Decision
Graphs of the dependence of the coordinates of bodies from time to time are shown in the picture.
Make a drawing based system of equations:
Deciding the system with respect to t C. We get:
Then the distance from the point A. to the meeting place:
.
Motor boat passes the distance between two points A. and B. for the river during the time t. 1 \u003d 3 h, and raft - during t. \u003d 12 h. How long t. 2 will spend the engine boat on the return path?
Decision
Let be s. - distance between points A. and B., v. - speed of boat relative to water, and u. - flow rate. Expressing the distance s. Three times - for the flesh, for a boat moving along the flow, and for a boat moving against the flow, we obtain the system of equations:
Solving the system, we get:
The Escalator Metro descends the man walking on it for 1 min. If a person is twice as fast, he will go down in 45 s. How long is the person standing on the escalator descends?
Decision
Denote by letter l. the length of the escalator; t. 1 - the time of descent of a person going at speeds v.; t. 2 - the time of descent of a person walking at speed 2 v.; t. - Time of descent standing on a person's escalator. Then, having calculated the length of the escalator for three different cases (the person comes with speed v., at speed 2 v. and stands on an escalator stationary), we obtain the system of equations:
Deciding this system of equations, we get:
Man runs through the escalator. For the first time he counted n. 1 \u003d 50 steps, a second time, moving to the same side with speed, three times more, it counted n. 2 \u003d 75 steps. How many steps did he count on a fixed escalator?
Decision
Since, with an increase in speed, a person counted more than a taper, which means the direction of the speeds of the escalator and the person coincide. Let be v. - human speed relative to the escalator, u. - Escalator speed, l. - the length of the escalator, n. - number of steps on a fixed escalator. The number of steps that fit in a unit of the length of the escalator is equal n./l.. Then the time of the person's stay on the escalator with its movement relative to the escalator at speeds v. equally l./(v.+u.), and the path passed on the escalator is equal to v.l./(v.+u.). Then the number of steps numbered on this path is equal. Similarly, for the case when human speed relative to escalator 3 v., I get.
Thus, we can make a system of equations:
Excluding attitude u./v.We will get:
Between two points located on the river at a distance s. \u003d 100 km one from the other, runs the boat, which, going downstream, passes this distance during the time t. 1 \u003d 4 h, and against the current, - during t. 2 \u003d 10 h. Determine the flow rate of the river u. And the speed of the boat v. Regarding water.
Decision
Expressing the distance s. Twice, - for a boat walking downstream, and a boat coming against the current - we obtain the system of equations:
Solving this system, we get v. \u003d 17.5 km / h, u. \u003d 7.5 km / h.
Plot passes by the pier. At this point in the village, located at a distance s. 1 \u003d 15 km from the pier, down the river goes the engine boat. She reached the village for the time t. \u003d 3/4 h and turning back, met the raft at a distance s. 2 \u003d 9 km from the village. What are the flow rate of the river and the speed of the boat relative to the water?
Decision
Let be v. - motor boat speed, u. - The flow rate of the river. Since from the moment the engine boat from the pier until the engine of the engine boat with a raft, obviously, will pass the same time for the fleet, and for a motor boat, then the following equation can be made:
where to the left is the expression of the time passed until the meeting, for the raft, and on the right - for a motor boat. We write the equation for the time that the engine boat spent on overcoming the path s. 1 from the pier to the village: t.=s. 1 /(v.+u.). Thus, we obtain a system of equations:
Where do you get v. \u003d 16 km / h, u. \u003d 4 km / h.
Column of troops during the hike moves at speed v. 1 \u003d 5 km / h, stretching on the road to distance l. \u003d 400 m. The commander, located in the tail of the column, sends a biker with the instruction of the head detachment. The cyclist goes and rides at speeds. v. 2 \u003d 25 km / h and, on the go, by following the order, immediately returns back at the same speed. After how much time t. After receiving the order, he returned back?
Decision
In the reference system associated with the column, the speed of the cyclist when moving to the head squad is equal to v. 2 -v. 1, and when moving back v. 2 +v. one . Therefore:
Simplifying and substituting numerical values, we get:
.
Wagon wagon d. \u003d 2.4 m moving at speeds v. \u003d 15 m / s, a bullet was broken, flying perpendicular to the movement of the car. Displacement of the holes in the walls of the car relative to each other equals l. \u003d 6 cm. What is the speed of the bullet?
Decision
Denote by letter u. bullet speed. The flight time of the bullet from the wall to the Wagon wall is equal to the time for which the car passes the distance l.. Thus, it is possible to make an equation:
From here to find u.:
.
What is the speed of droplets v. 2 steeply falling rain if the chauffeur of a car noticed that raindrops do not leave the trace on the rear window, tilted forward at the angle α \u003d 60 ° to the horizon when the vehicle speed v. 1 More than 30 km / h?
Decision
As can be seen from the drawing,
to drop raindrops left a trace on the rear window, come out that the time of passing the distance drops h. It was equal to the time for which the car will pass distance l.:
Or expressing from here v. 2:
It is raining outside. In which case, a bucket that stands in the body of a truck will be filled faster water: When is the car moves or when it costs?
Answer
Equally.
At what speed v. And at what course the plane should fly to take t. \u003d 2 hours fly exactly to the north way s. \u003d 300 km, if the northwest wind blows at the angle α \u003d 30 ° to meridian at speeds u. \u003d 27 km / h?
Decision
We write the system of equations in the figure.
Since the plane must fly strictly to the north, the projection of its speed on the axis Oy. v. Y is equal y.-Noting wind speed u. y.
Deciding this system, we find that the plane must keep the course for the north-west at an angle of 4 ° 27 "to the meridian, and its speed should be 174 km / h.
On the smooth horizontal table moves at speed v. Black board. What form track will leave on this chalkboard chalk, abandoned horizontally at speed u. Perpendicular to the direction of the board movement, if: a) friction between chalk and the board is negligible; b) friction is great?
Decision
The chalk will leave a trace on the board, which is a straight line, the angle of ArctG ( u./v.) With the direction of the motion of the board, i.e. coincides with the direction of the sum of the velocity of the boards and chalk. This is also true for the case of a) and for the case b), since the friction force does not affect the direction of movement of the chalk, since it lies on one straight line with a velocity vector, it only reduces the rate of chalk, so the trajectory in the case of b) may not reach the edge of the board.
Ship comes out of item A. And goes with speed v.constituting corner α With line AB.
At what angle β To line AB should be released from the point B. Torpider so that she struck the ship? Torpée need to be released at the moment when the ship was in paragraph A.. Torpeda speed is equal u..
Decision
Point C. The picture is a place of meeting the ship and torpedoes.
AC = vt., BC. = uT.where t. - Time from the start until the meeting. According to the Sinusov Theorem
From here to find β :
.
To the slider that can move along the guide rail,
the cord is attached, which has been shifted through the ring. Cord choose at speed v.. At what speed u. Moves the slider at the time when the cord is from the corner guide α ?
Answer and solution
u. = v./ COS. α.
For a very short period of time Δt. The slider moves to the distance AB = ΔL..
The cord for the same time is chosen for the length AC = ΔL.cos. α (angle ∠ ACB. can be considered direct because the angle Δα very small). Therefore, you can write: ΔL./u. = ΔL.cos. α /v.From! u. = v./ COS. α , which means that the speed of choosing a rope is equal to the projection of the speed of the slider to the direction of the rope.
Workers raising cargo
pull the ropes at the same speed v.. What speed u. He 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 the speed of cargo u. On the direction of the rope is equal to the speed of the rope v. (see task 15), i.e.
u.cos. α = v.,
u. = v./ COS. α.
Rod length l. \u003d 1 m hinged with couplings A. and B.which move along two mutually perpendicular to the raids.
Coupling A. moves at a constant speed v. A \u003d 30 cm / s. Find speed v. B couplings B. at the moment when the angle OAB \u003d 60 °. Accepting the beginning of the time the moment when the coupling A. was at point O., Determine the distance OB. and speed of the coupling B. In the function of time.
Answer and solution
v. B \u003d. v. A CTG. α
\u003d 17.3 cm / s; . 
.
At any time of the projection of speeds v. A I. v. B ends of the rod
the rod axis is equal to each other, since otherwise the rod would have to shorten or lengthen. So you can write: v A.cos. α = v B.sin. α . From v B. = v A.cTG. α .
At any time for a triangle OAB Fair the theorem Pythagora: l. 2 = Oa. 2 (t.) + OB. 2 (t.). Find out here OB.(t.):. Insofar as Oa.(t.) = v A T.then finally write the expression for OB.(t.) So: .
Since CTG. α
at any time is equal Oa.(t.)/ OB.(t.), then you can record an expression for dependence v B. from time: .
The tank is moving at a speed of 72 km / h. What speed is moving relative to the earth: a) the upper part of the caterpillar; b) the lower part of the caterpillar; c) the point of the caterpillar that this moment Moves vertically to the tank?
Answer and solution
a) 40 m / s; b) 0 m / s; c) ≈28.2 m / s.
Let be v. - Speed \u200b\u200bspeed tank relative to the Earth. Then the speed of any point of the caterpillar relative to the tank is also equal v.. The speed of any point of the caterpillar relative to the Earth is the sum of the velocity vectors of the tank relative to the Earth and the speed of the caterpillar point relative to the tank. Then for the case a) 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 speeds v. 1 \u003d 40 km / h, second - at speed v. 2 \u003d 60 km / h. To find middle speed On the distance traveled.
2. The car drove half the way at speeds v. 1 \u003d 60 km / h, the remaining part of the path it was half time at speeds v. 2 \u003d 15 km / h, and the last plot - at speeds v. 3 \u003d 45 km / h. Find the average vehicle speed all over the way.
Answer and solution
1. v. cf \u003d 48 km / h; 2. v. CP \u003d 40 km / h.
1. Let s. - all the way t. - Time spent on overcoming the entire path. Then the average speed throughout the path is equal s./t.. Time t. It consists of the amount of time intervals spent on overcoming the 1st and 2nd half of the way:
Substitting this time to an expression for medium speed, we get:
.(1)
2. The solution to this task can be reduced to the solution (1.), if you first determine the average speed in the second half of the path. Denote this speed v. CP2, then you can write:
where t. 2 - time spent on overcoming the 2nd half of the path. The path passed during this time consists of a path traveled at speeds. v. 2, and the path passed at speeds v. 3:
Substituting it in the expression for v. CP2, we get:
.
.
Train the first half of the way was at speed in n.\u003d 1.5 times greater than the second half of the way. Average train speed all over the way v. CP \u003d 43.2 km / h. What are the train speeds on the first ( v. 1) and the second ( v. 2) Halfs?
Answer and solution
v. 1 \u003d 54 km / h, v. 2 \u003d 36 km / h.
Let be t. 1 I. t. 2 - time passing by train, respectively, the first and second half of the way, s. - All the way traveled by train.
We will make a system of equations - the first equation is an expression for the first half of the way, the second - for the second half of the path, and the third - for the entire path traveled by the train:
Making a substitution v. 1 =nV 2 and solving the resulting system of equations, we get v. 2 .
Two balls began simultaneously and with the same speed to move along the surfaces with the form shown in the figure.
How will the speed and times of the balls of the balls be distinguished by the time they arrive at the point B.? Friction neglect.
Answer and solution
Speeds will be the same. The time of movement of the first ball will be greater.
The figure shows approximate charts of the movement of balls.
Because The paths passed by balls are equal, then the area of \u200b\u200bthe shaded figures is also equal (the area of \u200b\u200bthe 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 clause B. and returns back to item A.. The speed of the aircraft in the windless weather is equal v.. Find the ratio of medium speeds of the entire flight for two cases, when the wind blows during the flight: a) along the line AB; b) perpendicular lines AB. Wind speed is equal u..
Answer and solution
Airplane flight time from point A. to clause B. And back in the case when the wind blows along the line AB:
.
Then the average speed in this case:
.
In case the wind blows perpendicularly lines AB, the aircraft speed vector should be directed at an angle to the line AB So, to compensate the effect of wind:
The flight time "Round-back" in this case will be:
Flight speed of the aircraft to the point B. And back the same and equal:
.
Now you can find the ratio of average speeds obtained for the considered cases:
.
Distance between two stations s. \u003d 3 km Metro train passes with an average speed v. CP \u003d 54 km / h. At the same time, he spends time on acceleration t. 1 \u003d 20 s, then goes uniform some time t. 2 and to slow down until the full stop spends time t. 3 \u003d 10 s. Build a train speed chart and determine the highest train speed v. Max.
Answer and solution
The figure shows a schedule of the speed of the train.
The path traveled by the train is numerically equal to the area of \u200b\u200bthe figure, limited by the schedule and the time axis t.Therefore, you can write the system of equations:
From the first equation express t. 2:
then from the second equation of the system we find v. Max:
.
From the moving train dismissed the last car. The train continues to move at the same speed v. 0. How will the paths traveled by train and the car at the time of stopping the car? It is considered that the car moves equifiable. Solve the task also graphically.
Answer
At the moment when the train was moved, seeking began to run evenly along the train at speeds v. 0 \u003d 3.5 m / s. Taking train movement is equidalized, determine the speed of the train v. At that moment, when the accompanying is becoming steady.
Answer
v.\u003d 7 m / s.
The graph of the speed of the speed of some body from time to time is shown in the figure.
Draw graphs of the dependence of the acceleration and the coordinates of the body, as well as the path passed by them.
Answer
The graphs of the dependence of the acceleration, the coordinates of the body, and the path passed on them from time are shown in the figure.
The chart of the body acceleration from time to time has the form shown in the figure.
Draw graphs of the dependence of the speed, offset and path traveled by the body, from time to time. The initial body velocity is zero (on the spanning section, the acceleration is zero).
The body begins to move from the point. A. with speed v. 0 and after a while it gets to the point B..
What path the body passed if it moved equally to the acceleration, numerically equal a.? Distance between points A. and B. equally l.. Find the middle body speed.
The figure shows a graph of the body coordinate of the body.
After moment t.=t. 1 Curve graphics - Parabola. What movement is shown on this schedule? Build a chart of body speed dependence on time.
Decision
On the plot from 0 to t. 1: Uniform movement at speeds v. 1 \u003d TG. α ;
on the plot OT. t. 1 BE t. 2: Equalized movement;
on the plot OT. t. 2 BE t. 3: Equal asked movement in the opposite direction.
The figure shows a chart of body speed dependence on time.
The figure shows velocity graphics for two points moving on one straight line from the same initial position.
Known moments of time t. 1 I. t. 2. At what point t. 3 points will meet? Build traffic schedules.
For which second, from the beginning of the movement, the path passed by the body in equal asked movement, three times more paths passed in the previous second if the movement occurs without the initial speed?
Answer and solution
For the second second.
The easiest way to solve this task graphically. Because The path passed is numerically equal to the area of \u200b\u200bthe figure under the speed line of the speed of the speed, then from the figure it is obvious that the path passed in the second second is equal to the area of \u200b\u200bthree triangles), 3 times the path traveled for the first second (the area is equal to Square One triangle).
The trolley should carry the load in the shortest time from one place to another at a distance L.. It can accelerate or slow down its movement only with the same largest and constant acceleration. a., pass on to a uniform movement or stopping. What the greatest speed v. Should the trolley reach to fulfill the above requirement?
Answer and solution
Obviously, the car transports the cargo in the minimum time if it is the first half of the way to move with acceleration + a., and the remaining half with acceleration - a..
Then you can write down the following expressions: L. = ½· vt. 1 ; v. = ½· aT. 1 ,
where do you find the maximum speed:
The jet plane flies at speeds v. 0 \u003d 720 km / h. From some moment on, the plane is moving with acceleration for t.\u003d 10 s and in the last second goes s.\u003d 295 m. Determine Acceleration a. and final speed v. plane.
Answer and solution
a.\u003d 10 m / s 2, v.\u003d 300 m / s.
I will shown the aircraft speed schedule in the figure.
Aircraft speed at time t. 1 equal v. 1 = v. 0 + a.(t. 1 - t. 0). Then the way passed by the plane during t. 1 BE t. 2 Ravenna s. = v. 1 (t. 2 - t. 1) + a.(t. 2 - t. 1) / 2. From here you can express the desired amount of acceleration a. and, substituting the values \u200b\u200bfrom the condition of the problem ( t. 1 - t. 0 \u003d 9 s; t. 2 - t. 1 \u003d 1 s; v. 0 \u003d 200 m / s; s. \u003d 295 m), we get acceleration a. \u003d 10 m / s 2. The final speed of the aircraft v. = v. 2 = v. 0 + a.(t. 2 - t. 0) \u003d 300 m / s.
The first train car passed by the observer standing on the platform, for t. 1 \u003d 1 s, and the second - for t. 2 \u003d 1.5 s. Vagon length l.\u003d 12 m. Find Acceleration a. Train and its speed v. 0 at the beginning of the observation. Train movement is considered equalized.
Answer and solution
a.\u003d 3.2 m / s 2, v. 0 ≈13.6 m / s.
The path traveled by train by the time t. 1 equal:
and the path to the moment of time t. 1 + t. 2:
.
From the first equation we find v. 0:
.
Substitting the resulting expression in the second equation, we get acceleration a.:
.
Ball, fastened up inclined plane, passes successively two equal length length l. Each continues to move on. The first segment of the ball was held for t. Seconds, the second - for 3 t. seconds. Find speed v. Ball at the end of the first cutting path.
Answer and solution
Since the ball in question is reversibly reversible, it is advisable to choose the beginning of the countdown of two segments. In this case, the acceleration when moving on the first segment will be positive, and when moving on the second segment is negative. Initial speed in both cases is equal v.. Now write a system of equation equations for the paths passed by the ball:
Exclude acceleration a., we get the desired speed v.:
The board, divided into five equal segments, begins to slide along the inclined plane. The first segment passed past the mark made on the inclined plane in the place where the front edge of the board was at the beginning of the movement, for τ \u003d 2 s. For some time passes Past of this mark the last segment of the board? The movement of the board is considered equivalent.
Answer and solution
τ n \u003d 0.48 s.
Find the length of the first segment:
Now write the equations of motion for start points (time t. 1) and the end (time t. 2) Fifth segment:
Following the substitution found above the length of the first segment instead l. and finding a difference ( t. 2 - t. 1), we get the answer.
The bullet, flying at a speed of 400 m / s, hit in an earthen shaft and penetrates it to a depth of 36 cm. How much time did it move inside the shaft? What acceleration? What was her speed at a depth of 18 cm? At what depth the speed of the bullet decreased three times? The movement is considered equalized. What will the rate of bullets be equal to the time when the bullet goes 99% of its path?
Answer and solution
t. \u003d 1.8 · 10 -3 C; a. ≈ 2.21 · 10 5 m / s 2; v. ≈ 282 m / s; s. \u003d 32 cm; v. 1 \u003d 40 m / s.
The time of movement of the bullet inside the shaft will be found from the formula h. = vt./ 2, where h. - Full depth of immersion bullets, from where t. = 2h./v.. Acceleration a. = v./t..
On the inclined board, let me roll up the ball. On distance l. \u003d 30 cm from the beginning of the path the ball visited twice: through t. 1 \u003d 1 s and through t. 2 \u003d 2 s after the start of the movement. Determine the initial speed v 0 and acceleration a. The movement of the ball, considering it constant.
Answer and solution
v. 0 \u003d 0.45 m / s; a. \u003d 0.3 m / s 2.
The dependence of the speed of the ball on time is expressed by the formula v. = v. 0 - aT.. At the time of time t. = t. 1 I. t. = t. 2 The ball had the same largest and opposite in the direction of speed: v. 1 = - v. 2. But v. 1 = V. 0 - aT. 1 I. 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 equally, then the distance l. It can be expressed as follows:
Now you can create a system of two equations:
,
deciding which, we get:
The body falls from a height of 100 m without initial speed. How long does the body go through the first and last meters of its path? What path is the body for the first, over the last second of your movement?
Answer
t. 1 ≈ 0.45 s; t. 2 ≈ 0.023 s; s. 1 ≈ 4.9 m; s. 2 ≈ 40 m.
Determine the open position of the photographic shutter τ if when photographing a ball falling along the vertical centimeter scale from the zero marker without initial speed, a strip extended from the negative was obtained n. 1 BE n. 2 divisions of the scale?
Answer
.
A free falling body has passed the last 30 m during 0.5 s. Find the height of the fall.
Answer
A free falling body over the last second of the fall passed 1/3 of its way. Find the time of the fall and the height from which the body fell.
Answer
t. ≈ 5.45 s; h. ≈ 145 m.
What initial speed v. 0 you need to throw down the ball from the height h.so that he jumped to height 2 h.? Friction about the air and other losses of mechanical energy neglected.
Answer
At which time, two drops were taken away from the roof cornice, if two seconds after the beginning of the fall, the second drops between the drops was 25 m? Friction about the air neglected.
Answer
τ ≈ 1 s.
The body throws up vertically. The observer notices the time interval t. 0 between two moments when the body passes the point B.inspected h.. Find the initial speed of cast v. 0 and time of all body movement t..
Answer
; .
From the point A. and B.vertical (point A. above) l. \u003d 100 m from each other, thrown at the same time two bodies at the same speed of 10 m / s: from A. - vertically down, from B. - vertically up. After how much time and in what place will they meet?
Answer
t. \u003d 5 s; 75 m below the point B..
The body is thrown vertically up at the initial speed v. 0. When it reached the highest point of the path, from the same starting point at the same speed v. 0 The second body is thrown. At what height h. From the initial item they will meet?
Answer
Two bodies thrown up vertically up from the same point with the same initial speed. v. 0 \u003d 19.6 m / s with a period of time τ \u003d 0.5 s. After what time t. After throwing the second body and at what height h. will there be bodies?
Answer
t. \u003d 1.75 s; h. ≈ 19.3 m.
Aerostat rises from ground vertically up with acceleration a. \u003d 2 m / s 2. Through τ \u003d 5 seconds the subject fell out of its movement. After how much time t. This subject will fall to the ground?
Answer
t. ≈ 3.4 s.
With aerostat descending at speed u.Throw up the body at speeds v. 0 relative to the Earth. What will be the distance l. Between the aerostat and body by the time of the highest lifting of the body relative to the Earth? What is the greatest distance l. Max between body and aerostat? After what time τ From the moment of throwing the body is painted with a balloon?
Answer
l. = v. 0 2 + 2uV 0 /(2g.);
l. max \u003d ( u. + v. 0) 2 /(2g.);
τ = 2(v. 0 + u.)/g..
The body at the point B. on high H. \u003d 45 m from the ground, begins to fall freely. At the same time from the point A.located at a distance h. \u003d 21 m below the point B., throw another body vertically up. Determine the initial speed v. 0 second body, if it is known that both bodies will fall on the ground at the same time. Air resistance to neglect. To accept g. \u003d 10 m / s 2.
Answer
v. 0 \u003d 7 m / s.
Body freely falls from height h.. At that time, another body was thrown from height H. (H. > h.) Vertically down. Both bodies fell to the ground at the same time. Determine the initial speed v. 0 second body. Check the correctness of the solution on the numerical example: h. \u003d 10 m, H. \u003d 20 m. Accept g. \u003d 10 m / s 2.
Answer
v. 0 ≈ 7 m / s.
The stone is thrown horizontally from the top of the mountain having a slope α. At what speed v. 0 should be thrown a stone so that it fell to the mountain at a distance L. From the top?
Answer
Two play the ball, throwing his friend. What is the greatest height reaches the ball during the game, if he flies 2 s from one player?
Answer
h. \u003d 4.9 m.
The plane flies at a constant height h. straight with speed v.. The pilot must reset the bomb to the target lying ahead of the plane. What an angle to the vertical should he see the goal at the time of resetting the bomb? What is at this point the distance from the target to the point over which the plane is located? Air resistance to the movement of the bomb not to take into account.
Answer
; .
Two bodies fall from the same height. On the way of one body, located at an angle of 45 ° to the horizon of the playground, from which this body is elastically reflected. How does the times and speed of the fall of these bodies differ?
Answer
The time of the body's fall, on the path of which was the platform, more, since the vector dialed by the speed changed its direction to the horizontal (with an elastic collision, the direction of the speed changes, but not its value), which means the vertical component of the velocity vector has become zero, at that time As another body, the velocity vector has not changed.
The rate of falling bodies are equal until the collision of one of the bodies with a platform.
The elevator rises with an acceleration of 2 m / s 2. At that moment, when its speed became 2.4 m / s, the bolt began to fall from the elevator ceiling. The elevator height is 2.47 m. Calculate the break time of the bolt and the distance passed by the bolt relative to the mine.
Answer
0.64 s; 0.52 m.
At some height at the same time, two bodies were thrown at the same point at an angle of 45 ° to vertical at a speed of 20 m / s: one down, another up. Determine the difference in heights Δh.which will be bodies through 2 p. How do these bodies move relative to each other?
Answer
Δ h. ≈ 56.4 m; Bodies are moving away from each other at a constant speed.
Prove that free movement There are relative speed of constant near the surface of the Earth.
From the point A. Body freely falls. At the same time from the point B. at an angle α The horizon throws another body so that both bodies face in the air.
Show that angle α does not depend on the initial speed v. 0 bodies abandoned from point B., and determine this angle if. Air resistance to neglect.
Answer
α \u003d 60 °.
The body is thrown at an angle α to the horizon at speed v. 0. Determine speed v. this body at height h. above the horizon. Does this speed depend on the challenge angle? The air resistance is not taken into account.
At an angle α \u003d 60 ° to the horizon thrown the body at the initial speed v.\u003d 20 m / s. After how much time t. It will move at an angle β \u003d 45 ° to the horizon? Friction is absent.
Of the three pipes located on Earth, the water jets are beaten with the same speed: at an angle of 60, 45 and 30 ° to the horizon. Find the relationship of the greatest heights h. Lifting water jets arising from each pipe and drop ranges l. Water to Earth. Air resistance to the movement of water jets not to take into account.
From the point lying on the upper end of the vertical diameter d. Some circumference, on the grooves installed along the different chord of this circle, are simultaneously starting to slide without friction.
Determine what time t. Loads reach a circle. How does this time depends on the angle of tilt chord to the vertical?
Initial speed of abandoned stone v. 0 \u003d 10 m / s, and later t.\u003d 0.5 s Stone Speed v.\u003d 7 m / s. What maximum height above the initial level is raised?
Answer
H. Max ≈ 2.8 m.
At some height, simultaneously from one point with the same speeds are thrown over all sorts of balls. What will be a geometric area of \u200b\u200bpoints of finding balls at any time? Air resistance to neglect.
Answer
The geometric location of points of finding balls at any time will be the sphere whose radius v. 0 t., and its center is located below the initial point by gT. 2 /2.
The goal located on the hill is visible from the place of the gun at an angle α To the horizon. Distance (horizontal distance from gun before the target) is equal L.. The target shooting is made at the angle of elevation β .
Determine the initial speed v. 0 shell falling into the target. The air resistance is not taken into account. With what corner of elevation β 0 Footing distance along the slope will be maximum?
Answer and solution
, .
Choose the coordinate system xoy. So that the reference point coincides with the gun. Now write the kinematic equations of the projectile movement:
Replace x. and y. on the goal coordinates ( x. = L., y. = L.tGα) and exclude t.We will get:
Range l. Extra flight along the slope l. = L./ COS. α . Therefore, the formula that we received can be rewritten:
,
this expression is maximally with the maximum value of the work.
therefore l. maximum at maximum value \u003d 1 or
For α \u003d 0 we get the answer β 0 = π / 4 \u003d 45 °.
Elastic body Falls from height h. On the inclined plane. Determine how much time t. After reflection, the body will fall on the inclined plane. How time depends on the angle of the inclined plane?
Answer
From the angle of the inclined plane does not depend.
From high H. on the inclined plane forming the horizon angle α \u003d 45 °, freely falls the ball and elastically reflected at the same speed. Find the distance from the place of the first blow to the second one, then from the second to the third, etc. Solve the task in general (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 time between the shot and its echo. What could be the error τ In determining the moments of the shot and arrival of echo, if the distance to the mountain is at least 1 km, and it needs to be determined with an accuracy of 3%? Sound speed in the air c.\u003d 330 m / s.
Answer
τ ≤ 0.09 p.
The depth of the well wants to measure with an accuracy of 5%, throwing a stone and noticing time τ through which a splash will be heard. Starting from what values τ Need to take into account the time of sounding sound? Sound speed in the air c.\u003d 330 m / s.
Answer
Most tasks on the movement of bodies with constant acceleration are mainly solved in the same way as the tasks for uniform straight traffic (See § 1.9). However, instead of one dependence of the coordinate from time to time, there will now be two: for coordinate and for the projection of speed depending on time:
2 "
X \u003d xq + v0xt +
2? Task 1.
Skater, disintegrating up to the speed V0 \u003d 6 m / s, began to slide equifiable. After a time t \u003d 30 with a speed skater module, moving straightforwardly, it became equal to V \u003d 3 m / s. Find the acceleration of the skater, considering it constant.
Decision. Compatible X axis with speed skate trajectory. For a positive axis direction, we select the direction of the initial velocity vector V0 (Fig. 1.66). As the skatehead moves with
standing acceleration, then vx \u003d V0x + AXT. Hence ah \u003d where
vX \u003d V and VQX \u003d V0, since vectors 50 and V have the same director
v - V0.
not as the X axis. Consequently, ah \u003d ---, ah \u003d -0.1 m / s2 and
a \u003d 0.1 m / s2. The "minus" sign indicates that the acceleration is directed opposite to the X axis.
Task 2.
Bruck on a smooth inclined plane reported the initial velocity V0 \u003d 0.4 m / s, directed upwards. The bar moves straightly with a constant acceleration, the module of which is a \u003d 0.2 m / s2. Find the velocities of the bar at the time of time equal to 1, 2, 3 s from the beginning of the movement. Determine the position of the bar at these points in time relative to the point where the bar had the speed and0. What is the path passed by Bruk for 3 s?
Decision. The acceleration of the bar is directed down along the plane both when it is lifted and when descent.
97
4-Myakyshev, 10 cl.
Compatible coordinate axis With the trajectory of movement. For the positive direction of the X axis, we will take the direction of the initial velocity vector and0. The beginning of the coordinates will choose at that point of the trajectory, where the bar had the speed V0 (Fig. 1.67).? The bar moves with a constant acceleration, so Vx \u003d VQX + AXT. Since V0x \u003d VQ, ah \u003d -a, then them \u003d V0 - AT. This formula is valid for any time.
We find projections and speed modules at the specified points in time:
vLX \u003d V0 - ATL \u003d 0.2 m / s, VX \u003d | ULJT | \u003d 0.2 m / s;
v2x \u003d V0- AT2 \u003d 0, V2 \u003d 0;
v3X \u003d V0 - AT3 \u003d -0.2 m / s, V3 \u003d | U3J \u003d 0.2 m / s.
Since VLX\u003e 0, then the speed is directed to the same side as the X axis. The "minus" sign in the projection V3x indicates that the speed V3 is directed towards opposite X axis. So it should be, because after stopping ( V2 \u003d 0) The bar will start sliding down the plane.
We will find the position of the bar for the specified points of time:
.2
aT \\ _. 0.2 m _ 0 x1 \u003d v0t1 - \u003d 0.4 m - - \u003d 0.3 m,
.2 AT2.
x2 \u003d v0t2 - -g- \u003d 0.8 m - 0.4 m \u003d 0.4 m,
.2 AT3.
x3 \u003d v0t3 - -g- \u003d 1.2 m - 0.9 m \u003d 0.3 m.
Note that at a point in the coordinate of 0.3 m (LG1 \u003d LG3) (see Fig. 1.67) the body was twice (when climbing and descent). At the same time, the body had velocity equal to the module (L\u003e 1 \u003d L\u003e 3), but opposite in the direction: v1 - -v3.
At point A with coordinate X2 (see Fig. 1.67) Speed \u200b\u200bv2 \u003d 0. There was a change in the direction of speed. At the time of time T3 \u003d 3, the bar was located at point B with the coordinate x3. Consequently, the bare path passed
s - OA + AB \u003d 2x2 - x3 \u003d 0.5 m.
Task 3.
Figure 1.68, and shows a graph of the point of projection of the point of time. Build a graph of the dependence of the coordinates from time to time, if the initial coordinate і, \u003d 5 m, build a graph of the track of the distance from time to time.
Decision. First we construct a graph of the dependence of the coordinates from time to time. The first 2 with the point moved equifiable oppositely axis x (VLX in the following 2 with the movement was equal to the same direction as first (V2X
Y t with
From 4 to 6 with the point again moved equifiable in the same direction, therefore x3 \u003d x2 + lh3 \u003d -1 m - 3 m \u003d -4 m. Grabol - Parabola DL - its top.
8 ya t, with
From 6 to 8 with the point moved equally in the positive direction of the X axis (V4x\u003e 0). Graph - Parabola dxej. By the end of the 8th second of the coordinate ї4 \u003d -4m + zm \u003d -1 M. The point further moved equible in the same direction (V5x\u003e 0): \u003d -1 M + 3 m \u003d 2 m. Chart - Parabola E1FV? 1. When building a schedule, it is necessary to consider that the path is a non-negative value and cannot be reduced in
movement process.
The chart consists of segments Parabola A2B2, B2C2, C2D2, D2E2, E2F2 (Fig. 1.68, B).
Exercise 3.
A small cube on a smooth inclined plane reported the initial speed I0 \u003d 8 m / s, directed upwards. The cube moves straightly with a constant acceleration, the module of which is a \u003d 2 m / s2. Find the position of the cube with respect to the point of the plane, where the cube is reported speed V0, at the time of time 2, 4, 6 s from the beginning of the movement, as well as the speed of the cube at the same time points. What is the path passed by a cube for 5 s?
Two cyclists travel towards each other. One of them at the initial speed of 18 km / h rises to the mountain equible with a constant acceleration, the module of which is 20 cm / C2. Another cyclist with an initial speed of 5.4 km / h is descended from the mountain with the same by the acceleration module. What time will they meet? At what distance from the foot of the mountain there will be a meeting and what path will each of them go to this point? The distance between the cyclists at the initial moment was 195 m.
Figure 1.69 shows graphs I, II and III projections of the speed of three bodies moving straightforwardly. Describe the features of the movement of tel. What does the point and the intersection of graphs correspond? Find body acceleration modules. Record the formula to calculate the speed of each body.
The distance 20 km between the two stations, the train takes place at a speed, the average module of which is 72 km / h, and he spends 2 minutes to accelerate, and then comes at a constant speed. To braking until a complete stop, the train spends 3 min. Determine the maximum train speed module.
Sledge, rolling from the mountain, in the first 3 C pass 2 m, and in the next 3 C - 4 m. Considering the movement is equivalent, find the acceleration module and the initial scene module.
The body moving is equilibly with the initial speed of 1 m / s, acquires, having passed a certain distance, the speed is 7 m / s. What was the body speed in the middle of this distance? VX, m / s
vx\u003e m / s s
-4"
Fig. 1.70
4
ABOUT
Fig. 1.69
t, with a straight line starts to move a constant acceleration point. After T1, after the start of its movement, the direction of acceleration point changes to the opposite, remaining unchanged by module. Determine what time T2 after the start of the movement
"The point will return to its original position.
The trolley should carry the load as soon as possible from one place to another, removed from the first to distance L. It can increase or decrease its speed only with the same acceleration module equal to a. In addition, it can move at a constant speed. What is the highest velocity module should the trolley be achieved to make the above condition?
Figure 1.70 shows a graph of the dependence of the projection of the speed of the point moving straightly, from time to time. Build a graph of the dependence of the coordinates from time to time, if \u003d 4.5 m. Build a path of dependence on time.
1. The body moves with constant acceleration and zero initial speed. Show graphically that the ways passable by the body for successive equal intervals are related as consecutive odd numbers.
Decision . With an equilibrium body movement with a zero initial speed, its speed over time t.varies according to the law
where a.- Acceleration.
We construct a speed chart (see Fig.) And we note on the axis t.equal intervals OA 1 =BUT 1 BUT 2 =BUT 2 BUT 3 =BUT 3 BUT 4 \u003d ...; From the point BUT 1 ,BUT 2, ... We carry out the dotted line of vertical straight to the intersection with speed schedule at points IN 1 ,IN 2 ,IN 3, .... Then the path passed during the first gap is numerically equal to the triangle area OA 1 IN one ; Ways passed over the subsequent gaps are equal to the areas of the respective trapezium. From the graph it is seen that the area of \u200b\u200bthe first trapezium BUT 1 BUT 2 IN 2 IN 1 is three triangle squares OA 1 IN one ; The area of \u200b\u200bthe next trapezium BUT 2 BUT 3 IN 3 IN 2 is equal to five triangle squares OA 1 IN 1, etc. Consequently, the ratio of paths passed by the body for successive equal periods of time is:
S. 1:S. 2:S. 3: …: S. n. = 1:3:5: …: (2n. – 1).
2. In the fifth second of the equilibrium movement with zero initial speed, the body passes the way S. 2 \u003d 36 m. What way S. 1 passes the body for the first second of this movement?
Decision . From the decision of the previous task it follows that
S. 1:S. 5 = 1:9.
Hence,
4 m.
3. A free falling body over the last second of the fall passed 1/3 of its way. Finding time t. and height h.With which body fell.
Decision . From the laws of the body movement with constant acceleration and zero initial speed, we obtain the following equations:
Here \u003d 1 s. Solving the resulting system of equations, we find:
Under the condition of the task t.\u003e 1. This condition satisfies the root 
5.4 s. Next, we get:
4. The balloon rises from the ground vertically up with acceleration a \u003d.2 m / s 2. After \u003d 10 ° C after the start of the movement from the ball basket, the subject was broken off. What maximum height h. m. will this subject rise? After what time t. 1 And at what speed V 1 will it fall to the ground?
R 
messing
.
The subject broke away from the basket of the balloon at the height 
having speed V 0 \u003d but, directed vertically upwards. Choose a reference system - axis OHdirected vertically upwards and depicting the position of the subject at the time of separation from the basket. Maximum height is equal
h. m. =h. 0 +S. m. ,
where 
- The path passed by the subject for the time after the separation to the rise to the maximum height, i.e.
It is further obvious that after the separation, the subject moves up during the time 
before stopping in the highest point, after which it falls freely from the height h. m. ; At the same time the time of its fall t.Naya from the relationship 
those. 
Hence,
The velocity of the subject that fell on the ground, we define the ratio
5. At which time, two drops of water were taken away from the roof cornice, if two seconds after the beginning of the fall of the second drop, the distance between them was S.\u003d 25 m?
Decision . Let - the time interval between the separation of the first and second drops, t. \u003d 2 s - time from the moment of separation of the second drop. Then by the time the second drop is the first drop of the first drop S. 0 = g. 2/2 and had a speed v 0 \u003d g.. Further obviously, the distance between the drops is equal
where 
- The path passed the first drop during t.,
- The path passed the second drop during the same time.
Hence,
Solving the resulting equation and taking into account that \u003e 0, we will find:
6. On the inclined board, let me roll up the ball. On distance l.\u003d 30 cm from the beginning of the throw the ball visited twice: through t. 1 \u003d 1 s and through t. 2 \u003d 2 s after the start of the movement. Determine the initial speed v 0 and acceleration a.ball, considering it constant.
Decision . We write the law of the ball of the ball by choosing the axis OX.directed along the movement of the ball:
I rewrite this equation so:
For x.=l.this equation has a root t. 1 I. t. 2 .
Therefore, by the Viitta theorem
Solving this system, we will find:
\u003d 30 cm / c 2,
\u003d 45 cm / s.
Comment
. This task can be solved otherwise, namely: taking advantage of the law of movement 
write two equations x.(t. 1) =l.and x.(t. 2) =l., and then solve the resulting system of equations with two unknown V 0 and a..
