Question

(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:

Part (a)

(i) From 0.0 s to 20.0 s

Step1: Analyze velocity - time graph

The velocity - time graph shows a horizontal line from \(t = 0.0\ s\) to \(t=20.0\ s\). A horizontal line on a velocity - time graph means that the velocity is constant. The velocity value from the graph is \(v = 2.0\ m/s\) (east direction as given in the graph). So the object is moving with a constant velocity of \(2.0\ m/s\) towards the east.

Step2: Conclusion

The object is moving with uniform motion (constant velocity) of \(2.0\ m/s\) east from \(0.0\ s\) to \(20.0\ s\).

(ii) From 20.0 s to 35.0 s

Step1: Analyze the slope of the graph

The velocity - time graph from \(t = 20.0\ s\) to \(t = 35.0\ s\) is a straight line with a positive slope. The initial velocity at \(t = 20.0\ s\) is \(v_i=2.0\ m/s\) and the final velocity at \(t = 35.0\ s\) is \(v_f = 5.0\ m/s\). The formula for acceleration \(a=\frac{v_f - v_i}{t_f - t_i}\). Here, \(t_f - t_i=35.0 - 20.0=15.0\ s\), \(v_f - v_i=5.0 - 2.0 = 3.0\ m/s\). So \(a=\frac{3.0}{15.0}=0.2\ m/s^2\). A positive slope on a velocity - time graph means the object is accelerating (speeding up) while moving east.

Step2: Conclusion

The object is accelerating uniformly (since the graph is a straight line) from \(2.0\ m/s\) to \(5.0\ m/s\) east, with an acceleration of \(0.2\ m/s^2\) from \(20.0\ s\) to \(35.0\ s\).

Part (b)

Step1: Find velocity at \(t = 10.0\ s\)

From the velocity - time graph, the velocity is constant from \(0.0\ s\) to \(20.0\ s\) (horizontal line). So at \(t = 10.0\ s\) (which is within the interval \(0.0\ s\) to \(20.0\ s\)), the velocity is the same as the velocity in that interval.

Step2: Read the velocity value

The velocity in the interval \(0.0\ s\) to \(20.0\ s\) is \(2.0\ m/s\) (east). So the velocity at \(t = 10.0\ s\) is \(2.0\ m/s\) east.

Answer:

The velocity of the object at \(10.0\ s\) is \(\boldsymbol{2.0\ m/s}\) (east).

Part (c)