Building dependency graphs
Coordinates from time
with uniform motion
Problem 7.1. Three dependency graphs are given v x = v x(t) (Fig. 7.1). It is known that X(0) = 0. Build dependency graphs X = X(t).
Solution. Since all graphs are straight lines, movement along the axis X equally variable. Because v x increases, then a x > 0.
In case 1 v x(0) = 0 and X(0) = 0, so the dependence X = X(t) quite simple: X(t) = = . Because the a x> 0 schedule X(t) will be a parabola with a vertex at point 0, the branches of which are directed upward (Fig. 7.2).
In case 2 X(t) = υ 0 x t + is also the equation of a parabola. Let's 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 v x < 0, а после момента t 1 v x> 0. This means that until the moment t 1 body moved 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 the moment 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 tÎ uniformly accelerated motion along the axis X without initial speed.
2. When tÎ uniform movement along the axis X.
3. When tÎ 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.
Now let's build a graph X = X(t).
On site schedule X(t) is a parabola with its vertex at point 0. Meaning 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).
The movement in the area is uniform at 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 – path traveled in time (5 s – 2 s) = 3 s, i.e. s 25 = (2 m/s)×(3 s) = 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= –2 m/s 2< 0, поэтому графиком X(t) is a parabola whose branches are directed downward. The vertex 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 – path traveled in a period of time, s 56 = 1 m, therefore, X(6) = 8 m + 1 m = 9 m.
On the site coordinate X(t) decreases, X(7) = x(6) – s 67 where s 67 – path traveled in 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! Solve for yourself: A1 (b, c), B3, B4.
Rules for constructing graphs x = x(t)
according to schedules v x = v x(t)
1. It is necessary to break up the schedule υ x = υ x(t) into sections so that at each section the following condition is satisfied: a x= const.
2. Take into account that in those areas where a x= 0, graph x = x(t) is straight, 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 upward if a x> 0 and down if a x < 0; б) координата t at the vertices of the parabola is at the point at which υ x(t c) = 0.
4. Between plot sections x = x(t) there should be no kinks.
5. If the value of the coordinate at the moment is known t 1 x(t 1) = X 1, then the coordinate value 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 graph 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 constructed sequentially for each section, starting from the point t = t 0, line x = x(t) is always continuous, so each subsequent section begins at the point where the previous one ends.
Problem 7.3. According to this schedule υ x = υ x(t) (Fig. 7.9, A) build a graph x = x(t). It is known that X(0) = 1.5 m.
Solution
.
1. Schedule υ x = υ x(t) consists of two sections: , on which a x < 0 и , на котором a x > 0.
2. On the site schedule x = x(t) is a parabola whose branches are directed downward, 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 intersects the axis X at the point X= 1.5 m, since x(0) = 1.5 m according to the problem conditions (Fig. 7.9, b).
3. On the site according to the schedule x = x(t) is also a parabola, but with branches up, since a x> 0. Its vertex is at the point tв = 3с, since υ x(3) = 0.
Coordinate values X at times 2s, 3s, 4s it is easy to find:
X(2) = X(1) – s 12 = 2 m – 1.5 m;
X(3) = X(2) – s 23 = 1.5 m – 1 m;
X(4) = X(3) + s 34 = 1 m + 1.5 m.
Stop! Solve for yourself: A1 (a), B5 (d, f, g).
Problem 7.4. According to this schedule x = = x(t) build a graph 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 parabola);
in the moment t= 2 s υ x> 0, since X grows;
Problem 40762
A body without initial velocity falls into a mine 100 km deep. Draw a graph of instantaneous speed versus time. Estimate the maximum speed of body movement.
Problem 10986
The equation rectilinear movement has the form x = At+Bt 2, where A = 3 m/s, B = -0.25 m/s 2. Construct graphs of coordinates and paths versus time for a given movement.
Problem 40839
The body moves in the direction opposite to the X axis at a speed of 200 m/s. Plot a graph of V x (t). Find graphically the displacement of the body along the X axis during the first 4 s of movement.
Problem 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; distance traveled by the body in t = 4 seconds from the start of movement; speed and acceleration of the body after t = 4 seconds from the start of movement; average speed and the average acceleration during the last second of movement. Plot graphs of the speed and acceleration of the body in the time interval from 0 to 4 seconds.
Problem 12242
Using the given equation for the path traveled by the body s = 4 + 2t + 5t 2, construct a graph of the speed versus time for the first 3 s. 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. Using 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 of the point; 4) write a formula for the dependence of speed on time v = f(t); 5) plot the dependence of the coordinate on time x = f(t) and speed on 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. Using the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 of the point; 3) acceleration a of the point; 4) write a formula for the dependence of speed on time v = f(t); 5) plot the dependence of the coordinate on time x = f(t) and speed on time v = f(t) in the interval 0
Problem 15935
The equation of motion of a point has the form x = 2+5t. Using the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 of the point; 3) acceleration a of the point; 4) write a formula for the dependence of speed on time v = f(t); 5) plot the dependence of the coordinate on time x = f(t) and speed on time v = f(t) in the interval 0
Problem 15937
The equation of motion of a point has the form x = 400–0.6t. Using the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 of the point; 3) acceleration a of the point; 4) write a formula for the dependence of speed on time v = f(t); 5) plot the dependence of the coordinate on time x = f(t) and speed on time v = f(t) in the interval 0
Problem 15939
The equation of motion of a point has the form x = 2t–t 2. Using the equation, determine: 1) the x 0 coordinate of the point at the initial moment of time; 2) initial speed v 0 of the point; 3) acceleration a of the point; 4) write a formula for the dependence of speed on time v = f(t); 5) plot the dependence of the coordinate on time x = f(t) and speed on time v = f(t) in the interval 0
Problem 17199
IN electrical circuit with low active resistance, containing a capacitor with a capacitance of C = 0.2 μF and an inductance coil of L = 1 mH, the current strength at resonance changes according to the law I = 0.02sinωt. Find the instantaneous current value, as well as the instantaneous voltage values on the capacitor and coil after 1/3 of the period from the beginning of the oscillations. Construct graphs of current and voltage versus time.
Problem 19167
A capacitor with a capacity of 0.5 μF 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 the equation for the current in oscillatory circuit. How long will it take for the amplitude of the current to decrease by a factor of 4? Plot a graph of current versus time.
A cart weighing m 1 =210 kg with a person weighing m 2 =70 kg moves freely horizontally at a speed v 1 =3 m/s. The person jumps in the direction opposite to the movement of the cart. The speed of the cart becomes equal to u 1 =4 m/s. Find the horizontal component of the velocity u 2x of the person relative to the cart during the jump.
