Math / Data & Statistics

QuestionA velocity-time graph for an object moving along the $x$ -axis is shown in Figure Q3 (b). (i) Plot a graph of the acceleration versus time. (ii) Determine the average acceleration of the object in the time intervals $\mathrm{t}=5.0 \mathrm{~s}$ to $\mathrm{t}=15.0 \mathrm{~s}$ and $\mathrm{t}=0$ to $\mathrm{t}=20.0 \mathrm{~s}$ .
Fiaure Q3 (b)

Studdy Solution

STEP 1

1. The velocity-time graph is piecewise linear.
2. Acceleration is the derivative of velocity with respect to time.
3. Average acceleration is the change in velocity divided by the change in time.

STEP 2

1. Determine the acceleration for each segment of the velocity-time graph.
2. Plot the acceleration versus time graph.
3. Calculate the average acceleration for specified time intervals.

STEP 3

Determine the acceleration for each segment:
- From $t = 0$ to $t = 5$ seconds, the velocity is constant at $-8 \, \text{m/s}$ . Therefore, acceleration $a = 0 \, \text{m/s}^2$ .
- From $t = 5$ to $t = 15$ seconds, the velocity changes linearly from $-8 \, \text{m/s}$ to $8 \, \text{m/s}$ . The change in velocity $\Delta v = 8 - (-8) = 16 \, \text{m/s}$ , and the change in time $\Delta t = 15 - 5 = 10 \, \text{s}$ . Thus, acceleration $a = \frac{\Delta v}{\Delta t} = \frac{16}{10} = 1.6 \, \text{m/s}^2$ .
- From $t = 15$ to $t = 20$ seconds, the velocity is constant at $8 \, \text{m/s}$ . Therefore, acceleration $a = 0 \, \text{m/s}^2$ .

STEP 4

Plot the acceleration versus time graph:
- From $t = 0$ to $t = 5$ seconds, $a = 0 \, \text{m/s}^2$ . - From $t = 5$ to $t = 15$ seconds, $a = 1.6 \, \text{m/s}^2$ . - From $t = 15$ to $t = 20$ seconds, $a = 0 \, \text{m/s}^2$ .
The graph will have horizontal lines at these acceleration values over the respective time intervals.

STEP 5

Calculate the average acceleration for the interval $t = 5$ to $t = 15$ seconds:
- The change in velocity $\Delta v = 8 - (-8) = 16 \, \text{m/s}$ . - The change in time $\Delta t = 15 - 5 = 10 \, \text{s}$ . - Average acceleration $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{16}{10} = 1.6 \, \text{m/s}^2$ .

STEP 6

Calculate the average acceleration for the interval $t = 0$ to $t = 20$ seconds:
- The change in velocity $\Delta v = 8 - (-8) = 16 \, \text{m/s}$ . - The change in time $\Delta t = 20 - 0 = 20 \, \text{s}$ . - Average acceleration $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{16}{20} = 0.8 \, \text{m/s}^2$ .
The average accelerations are:
- From $t = 5$ to $t = 15$ seconds: $1.6 \, \text{m/s}^2$ . - From $t = 0$ to $t = 20$ seconds: $0.8 \, \text{m/s}^2$ .

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