Question:

A car is moving along a straight line with variable velocity. The velocity vs. time graph for the car's motion is shown below:

(a) Determine the acceleration of the car for each segment of the graph labeled A, B, C, and D.

(b) Calculate the total displacement of the car during the time interval shown in the graph.

(c) Calculate the distance traveled by the car during the time interval shown in the graph.

Useful information:

Acceleration ($a$) is the rate of change of velocity ($v$) with respect to time ($t$), and can be computed as:

$$a = \frac{{\Delta v}}{{\Delta t}}$$

Displacement ($\Delta x$) is the change in position of an object and can be calculated using the area under the velocity vs. time graph according to the equation:

$$\Delta x = \int_{t_1}^{t_2}v(t)dt$$

Velocity ($v$) is the rate of change of displacement ($x$) with respect to time ($t$), and can be calculated as:

$$v = \frac{{\Delta x}}{{\Delta t}}$$

Speed is the absolute value of velocity, indicating only the magnitude.

Answer:

(a) To determine the acceleration for each segment of the graph, we need to calculate the slope of the graph at each interval.

Segment A:

The velocity graph is a horizontal line, indicating constant velocity. The slope of a horizontal line is zero, meaning there is no acceleration.

Segment B:

To calculate the acceleration, we will use the slope formula:

$$a_B = \frac{{\Delta v_B}}{{\Delta t_B}}$$

We can choose any two points on the line to calculate the slope:

$$a_B = \frac{{(15 \, \text{m/s}) - (0 \, \text{m/s})}}{{(10 \, \text{s}) - (0 \, \text{s})}}$$

$$a_B = \frac{{15 \, \text{m/s}}}{{10 \, \text{s}}}$$

$$a_B = 1.5 \, \text{m/s}^2$$

Thus, the acceleration for segment B is 1.5 m/s².

Segment C:

Similarly, we can calculate the acceleration for segment C using the slope formula:

$$a_C = \frac{{\Delta v_C}}{{\Delta t_C}}$$

Choosing two points on the line gives:

$$a_C = \frac{{(0 \, \text{m/s}) - (-20 \, \text{m/s})}}{{(30 \, \text{s}) - (10 \, \text{s})}}$$

$$a_C = \frac{{- 20 \, \text{m/s}}}{{20 \, \text{s}}}$$

$$a_C = - 1 \, \text{m/s}^2$$

Hence, the acceleration for segment C is -1 m/s² (negative sign indicates deceleration).

Segment D:

The velocity graph is a horizontal line again, indicating constant velocity. Therefore, the acceleration is zero.

(b) To find the total displacement of the car during the time interval, we need to calculate the area under the velocity vs. time graph between the given points.

The total displacement is the sum of the displacements for each segment:

$$\Delta x_{\text{total}} = \Delta x_A + \Delta x_B + \Delta x_C + \Delta x_D$$

For segment A, the displacement is given by:

$$\Delta x_A = v_A \cdot \Delta t_A$$

$$\Delta x_A = 10 \, \text{m/s} \cdot 15 \, \text{s}$$

$$\Delta x_A = 150 \, \text{m}$$

For segment B, the displacement is the area of the triangle:

$$\Delta x_B = \frac{1}{2} \cdot v_B \cdot \Delta t_B$$

$$\Delta x_B = \frac{1}{2} \cdot 15 \, \text{m/s} \cdot 10 \, \text{s}$$

$$\Delta x_B = 75 \, \text{m}$$

For segment C, the displacement is the area of the trapezoid:

$$\Delta x_C = \left(\frac{1}{2} \cdot (v_B + v_C) \cdot \Delta t_C\right) + (v_C \cdot \Delta t_C)$$

$$\Delta x_C = \left(\frac{1}{2} \cdot (15 \, \text{m/s} + 0 \, \text{m/s}) \cdot 20 \, \text{s}\right) + (0 \, \text{m/s} \cdot 20 \, \text{s})$$

$$\Delta x_C = \left(\frac{1}{2} \cdot 15 \, \text{m/s} \cdot 20 \, \text{s}\right) + 0 \, \text{m}$$

$$\Delta x_C = 150 \, \text{m}$$

For segment D, the displacement can be calculated similarly as segment A:

$$\Delta x_D = v_D \cdot \Delta t_D$$

$$\Delta x_D = 5 \, \text{m/s} \cdot 10 \, \text{s}$$

$$\Delta x_D = 50 \, \text{m}$$

Now, we can calculate the total displacement:

$$\Delta x_{\text{total}} = 150 \, \text{m} + 75 \, \text{m} + 150 \, \text{m} + 50 \, \text{m}$$

$$\Delta x_{\text{total}} = 425 \, \text{m}$$

Thus, the total displacement of the car during the time interval is 425 meters.

(c) To find the distance traveled by the car during the time interval, we need to sum the lengths of each segment in the velocity graph:

Distance = Length of segment A + Length of segment B + Length of segment C + Length of segment D

The length of each segment can be calculated as:

Length of segment A = $\Delta t_A \cdot |v_A| = 15 \, \text{s} \cdot 10 \, \text{m/s} = 150 \, \text{m}$

Length of segment B = $\Delta t_B \cdot |v_B| = 10 \, \text{s} \cdot 15 \, \text{m/s} = 150 \, \text{m}$

Length of segment C = $\Delta t_C \cdot |v_C| = 20 \, \text{s} \cdot 15 \, \text{m/s} = 300 \, \text{m}$

Length of segment D = $\Delta t_D \cdot |v_D| = 10 \, \text{s} \cdot 5 \, \text{m/s} = 50 \, \text{m}$

Hence, the distance traveled by the car during the time interval shown in the graph is:

Distance = 150 m + 150 m + 300 m + 50 m = 650 meters.

In summary, the acceleration for each segment is: A - 0 m/s², B - 1.5 m/s², C - -1 m/s², D - 0 m/s². The total displacement of the car is 425 meters, and the distance traveled is 650 meters.