Question

The graph shown in Fig. 4.108 shows the velocity of a police officer motorcycle plotted as a function of time.
a. The instantaneous accelerations at $$t=3 s$$, at $$t=7 s$$, and at $$t = 11s$$ are ...
b. The distances covered by the officer in the first $$5s$$, first $$9s$$, and first $$13s$$ are ...

Answer 1

a. At $$t=3s$$, the graph is horizontal, so the acceleration is 0. From $$t=5s$$ to $$t=9s$$, the acceleration is constant (from the graph) and equal to $$\dfrac { 45-20 }{ 9-5 } =6.25{ ms }^{ -2 }$$

From $$t=9s$$ to $$t=13s$$, the acceleration is constant and equal to $$\dfrac { 0-45 }{ 13-9 } =-11.25{ ms }^{ -2 }$$.

b. In the first five seconds, the area under the graph is the area of the rectangle, i.e., $$20\times 5=100 m$$.

Between $$t=5s$$ and $$t=9s$$, the area under the trapezoid is $$\left(\dfrac{1}{2}\right)(45+20)(4)=130m$$ and so the total distance in the first $$9s$$ is $$100+130=230 m$$.

Between $$t=9s$$ and $$t=13s$$, the area under the triangle is $$\left(\dfrac{1}{2}\right)(45)(4)=90 m$$, and so the total distance in the first $$13s$$ is $$230+90=320m$$.