Question

5.
(a) describe the motion of the object (use words and numbers):
(i) from 0.0 s to 20.0 s (ii) from 20.0 s to 35.0 s
(b) what is the velocity of the object at 10.0 s?
(c) what is the acceleration of the object at 30.0 s? does this represent the maximum magnitude of acceleration of the object?
(d) what is the displacement of the object first 40.0 s?

Explanation:

Step1: Analyze motion from 0.0 s to 20.0 s

The velocity - time graph is a horizontal line. For a velocity - time graph, when the graph is horizontal, the velocity is constant. The value of velocity \(v = 2.0\ m/s\) (east). So, the object is moving at a constant velocity of \(2.0\ m/s\) east.

Step2: Analyze motion from 20.0 s to 35.0 s

The velocity - time graph is a straight - line with a positive slope. The formula for the slope of a velocity - time graph is \(a=\frac{\Delta v}{\Delta t}\). The initial velocity \(v_{i}=2.0\ m/s\) and the final velocity \(v_{f}=6.0\ m/s\), and \(\Delta t=35 - 20=15\ s\). The slope (acceleration) \(a=\frac{6 - 2}{15}=\frac{4}{15}\approx0.27\ m/s^{2}\). So, the object is accelerating at approximately \(0.27\ m/s^{2}\) east.

Step3: Find velocity at \(t = 10.0\ s\)

Since from \(t = 0\) to \(t = 20\ s\) the velocity is constant. Looking at the graph (horizontal line in this interval), at \(t = 10.0\ s\), \(v = 2.0\ m/s\) (east).

Step4: Find acceleration at \(t = 30.0\ s\)

From \(t = 20\ s\) to \(t = 35\ s\), the acceleration is constant (straight - line part of the graph). Using the formula \(a=\frac{\Delta v}{\Delta t}\), \(v_{i}=2.0\ m/s\), \(v_{f}=6.0\ m/s\), \(\Delta t = 15\ s\), \(a=\frac{6 - 2}{15}=\frac{4}{15}\approx0.27\ m/s^{2}\). To check if it is the maximum acceleration:
The acceleration from \(t = 20\ s\) to \(t = 35\ s\) is \(a=\frac{4}{15}\approx0.27\ m/s^{2}\). Before \(t = 20\ s\), \(a = 0\) (constant velocity). After \(t = 35\ s\), \(a = 0\) (constant velocity). So, the acceleration at \(t = 30\ s\) (\(a=\frac{4}{15}\approx0.27\ m/s^{2}\)) is the maximum.

Step5: Calculate displacement in \(t = 40\ s\)

The displacement \(s\) from a velocity - time graph is the area under the graph.
The area has three parts:

• A rectangle from \(t = 0\) to \(t = 20\ s\): \(A_{1}=v\times t=2\times20 = 40\ m\)
• A trapezoid from \(t = 20\ s\) to \(t = 35\ s\): \(A_{2}=\frac{(v_{1}+v_{2})}{2}\times\Delta t=\frac{(2 + 6)}{2}\times15=60\ m\)
• A rectangle from \(t = 35\ s\) to \(t = 40\ s\): \(A_{3}=v\times t=6\times5 = 30\ m\)

The total displacement \(s=A_{1}+A_{2}+A_{3}=40 + 60+30=130\ m\) (east)

Answer:

(a)(i) The object is moving at a constant velocity of \(2.0\ m/s\) east. (ii) The object is accelerating at approximately \(0.27\ m/s^{2}\) east.
(b) \(2.0\ m/s\) (east)
(c) \(a=\frac{4}{15}\approx0.27\ m/s^{2}\) (east), and yes, it is the maximum magnitude of acceleration.
(d) \(130\ m\) (east)