task 12745
The speed of sound in water is 1450 m/s. At what distance are the nearest points that oscillate in opposite phases if the oscillation frequency is 906 Hz?task 17410
Two particles move in opposite directions from each other with speed u = 0.6s and v = 0.5s. How fast are the particles moving away from each other?task 26261
Between points A and B, located on opposite banks of the river, a boat runs. At the same time, he is always on the straight line AB (see figure). Points A and B are at a distance s = 1200 m from each other. River velocity u = 1.9 m/s. The straight line AB makes an angle α = 60° with the direction of the river flow. With what speed v relative to the water and at what angles β 1 and β 2 to the straight line AB should the boat move in both directions in order to pass from A to B and back in time t = 5 min?task 40481
A tennis ball with a speed of 10 m/s, after hitting the racket, flew in the opposite direction with a speed of 8 m/s. The kinetic energy of the ball has changed by 5 J. Find the change in the momentum of the ball.task 40839
The body moves in the direction opposite to the X axis, with a speed of 200 m/s. Plot the V x (t) dependency graph. Find graphically the movement of the body along the X axis in the first 4 seconds of movement.Problem 40762
A body without initial velocity falls into a shaft 100 km deep. Plot a graph of instantaneous speed versus time. Estimate the maximum speed of the body.
Problem 10986
The equation of rectilinear motion has the form x \u003d At + Bt 2, where A \u003d 3 m / s, B \u003d -0.25 m / s 2. Build graphs of coordinates and paths versus time for a given movement.
Problem 40839
The body moves in the direction opposite to the X axis, with a speed of 200 m/s. Plot the V x (t) dependency graph. Find graphically the movement of the body along the X axis in the first 4 seconds of movement.
Task 26400
The dependence of the X coordinate on time t is determined by the equation X = –1 + 2t – 3t 2 + 3t 3 . Determine the dependence of speed and acceleration on time; the distance traveled by the body in t = 4 seconds from the start of the movement; speed and acceleration of the body after t = 4 seconds from the start of motion; average speed and average acceleration for the last second of movement. Plot the speed and acceleration curves of the body in the time interval from 0 to 4 seconds.
Problem 12242
According to the given equation of the path traveled by the body s = 4 + 2t + 5t 2, construct a graph of the speed versus time for the first 3s. Determine the distance traveled by the body during this time?
Problem 15931
The equation of motion of a point has the form x = –1.5t. According to the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 points; 3) acceleration a point; 4) write the formula for the dependence of speed on time v = f(t); 5) build a graph of coordinates versus time x = f(t) and velocity versus time v = f(t) in the interval 0
Problem 15933
The equation of motion of a point has the form x = 1–0.2t 2 . According to the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 point; 3) acceleration a point; 4) write the formula for the dependence of speed on time v = f(t); 5) build a graph of coordinates versus time x = f(t) and velocity versus time v = f(t) in the interval 0
Problem 15935
The equation of motion of a point has the form x = 2+5t. According to the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 points; 3) acceleration a point; 4) write the formula for the dependence of speed on time v = f(t); 5) build a graph of coordinates versus time x = f(t) and velocity versus time v = f(t) in the interval 0
Problem 15937
The equation of motion of a point has the form x = 400–0.6t. According to the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 points; 3) acceleration a point; 4) write the formula for the dependence of speed on time v = f(t); 5) build a graph of coordinates versus time x = f(t) and velocity versus time v = f(t) in the interval 0
Problem 15939
The equation of motion of a point has the form x = 2t–t 2 . According to the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 points; 3) acceleration a point; 4) write the formula for the dependence of speed on time v = f(t); 5) build a graph of coordinates versus time x = f(t) and velocity versus time v = f(t) in the interval 0
Problem 17199
In an electrical circuit with low active resistance, containing a capacitor with a capacitance C = 0.2 μF and an inductance coil L = 1 mH, the current strength at resonance changes according to the law I = 0.02sinωt. Find the instantaneous value of the current strength, as well as the instantaneous values of the voltage on the capacitor and coil after 1/3 of the period from the onset of oscillations. Construct graphs of current and voltage versus time.
Problem 19167
A 0.5 μF capacitor was charged to a voltage of 20 V and connected to a coil with an inductance of 0.65 H and a resistance of 46 ohms. Find an equation for the current strength in an oscillatory circuit. After how long will the amplitude of the current decrease by 4 times? Plot a graph of current versus time.
Building dependency graphs
Coordinates from time
in uniform motion
Problem 7.1. Three dependency graphs are given υ x = υ x(t) (Fig. 7.1). It is known that X(0) = 0. Plot dependencies X = X(t).
Solution. Since all graphs are straight lines, the movement along the axis X equally variable. Because υ x increases, then a x > 0.
In case 1 υ x(0) = 0 and X(0) = 0, so the dependence X = X(t) is quite simple: X(t) = = . Because the a x> 0 chart X(t) will be a parabola with a vertex at point 0, the branches of which are directed upwards (Fig. 7.2).
In case 2 X(t) = υ 0 x t + is also the equation of a parabola. Find out where the vertex of this parabola will be. In the moment t 1 (t 1 < 0) проекция скорости меняет свой знак: до момента t 1 υ x < 0, а после момента t 1 υ x> 0. This means that up to the moment t 1 the body was moving in the negative direction of the axis X, and after the moment t 1 - in the positive direction. That is, at the moment t 1 body committed turn. Therefore, until t 1 coordinate X(t) decreased, and after the moment t 1 x(t) became
Stop! Decide for yourself: A2, B1, B2.
Problem 7.2. According to this schedule υ x = υ x(t) (Fig. 7.5) build graphs a x(t) and X(t). Count X(0) = 0.
Solution.
1. When tО uniformly accelerated motion along the axis X no initial speed.
2. When tО uniform motion along the axis X.
3. When tО uniformly slow motion along the axis X. In the moment t= 6 s the body stops, while a x < 0.
4. When tÎ uniformly accelerated movement in the direction opposite to the direction of the axis X, a x < 0.
Location on a x= 1 m/s;
Location on a x = 0;
Location on
a x = 
–2m/s 2 .
Schedule a x(t) is shown in Figure 7.6.
Let's build a graph now X = X(t).
On the plot plot X(t) is a parabola with vertex at point 0. The value X(2) = s 02 is equal to the area under the graph υ x(t) on the site, i.e. s 02 = 2 m. Therefore, X(2) = 2 m (Fig. 7.7).
On the site, the movement is uniform with a constant speed of 2 m / s. dependency graph X(t) in this section is a straight line. Meaning X(5) = X(2) + s 25 where s 25 - the path traveled in time (5 s - 2 s) = 3 s, i.e. s 25 \u003d (2 m / s) × (3 s) \u003d 6 m. Therefore, X(5) = = 2 m + 6 m = 8 m (see Fig. 7.7).
Rice. 7.7 Fig. 7.8
Location on a x\u003d -2 m / s 2< 0, поэтому графиком X(t) is a parabola whose branches point downwards. The top of the parabola corresponds to the moment in time t= 6 s, since υ x= 0 at t= 6 s. Coordinate value X(6) = X(5) + s 56 where s 56 - the path traveled for a period of time, s 56 = 1 m, therefore, X(6) = 8 m + 1 m = 9m.
On site coordinate X(t) decreases, X(7) = x(6) – s 67 where s 67 - the path traveled for a period of time, s 67 = = 1 m, therefore, X(7) = 9 m - 1 m = 8 m.
Final Schedule x = x(t) is shown in Fig. 7.8.
Stop! Decide for yourself: A1 (b, c), B3, B4.
Graphing Rules x = x(t)
according to schedules υ x = υ x(t)
1. You need to break the schedule υ x = υ x(t) into segments so that the following condition is fulfilled on each segment: a x= const.
2. Take into account that in those areas where a x= 0, graph x = x(t) is a straight line, and where a x= const ¹ 0, graph x = x(t) is a parabola.
3. When constructing a parabola, take into account that: a) the branches of the parabola are directed upwards if a x> 0 and down if a x < 0; б) координата t to the vertex of the parabola is at the point where υ x(t c) = 0.
4. Between sections of the graph x = x(t) should not have breaks.
5. If the value of the coordinate at the moment is known t 1 x(t 1) = X 1 , then the value of the coordinate at the moment t 2 > t 1 is determined by the formula x(t 2) = X 1 + s + – s- , where s+ - area under the graph υ x = υ x(t), s-- area above the chart υ x = υ x(t) Location on [ t 1 , t 2 ], expressed in units of length, taking into account the scale.
6. Initial coordinate value X(t) must be specified in the problem statement.
7. The graph is built sequentially for each section, starting from the point t = t 0 , line x = x(t) is always continuous, so each next segment starts at the point where the previous one ends.
Problem 7.3. According to this schedule υ x = υ x(t) (Fig. 7.9, a) plot x = x(t). It is known that X(0) = 1.5 m.
Solution
.
1. Graph υ x = υ x(t) consists of two sections: , on which a x < 0 и , на котором a x > 0.
2. On site schedule x = x(t) is a parabola whose branches are directed downwards, since a x < 0. Координата вершины t in = 1 s, since υ x(1) = 0, X(1) = X(0) + s 01 = = 1.5 m + 2.0 m. The parabola crosses the axis X at the point X= 1.5 m, since x(0) = 1.5 m according to the condition of the problem (Fig. 7.9, b).
3. On site schedule x = x(t) is also a parabola, but branches up, since a x> 0. Its vertex is at the point t in \u003d 3s, since υ x(3) = 0.
Coordinate values X at times 2s, 3s, 4s it is easy to find:
X(2) = X(1) – s 12 \u003d 2 m - 1.5 m;
X(3) = X(2) – s 23 \u003d 1.5 m - 1 m;
X(4) = X(3) + s 34 = 1 m + 1.5 m.
Stop! Decide for yourself: A1 (a), B5 (e, f, g).
Problem 7.4. According to this schedule x = = x(t) plot υ x = υ x(t). Schedule x = x(t) consists of parts of two parabolas (Fig. 7.10, a).
Solution.
1. Note that at the moment t= 0 υ x < 0, так как X decreases;
in the moment t= 1 s υ x= 0 (vertex of the parabola);
in the moment t= 2 s υ x> 0 because X is growing;
