Answer
(b) Displacement $=-1 \mathrm{m}$ and $\mathrm{Distance}=3 \mathrm{m}$
(a) Displacement $=1 \mathrm{m}$ and $\mathrm{Distance}=1 \mathrm{m}$
Work Step by Step
You have $\sin t=v(t), 0 \leq t \leq \frac{\pi}{2}$
Deplacement
$s(t)=\int_{0}^{\frac{\pi}{2}} \sin t d t=-\left.\cos t\right|_{0} ^{\frac{\pi}{2}}=-\left(\cos \frac{\pi}{2}-\cos 0\right)=1 m$
Distance
$: \int_{0}^{\frac{\pi}{2}}|\sin t| d t=\int_{0}^{\frac{\pi}{2}} \sin t d t=-\left.\cos t\right|_{0} ^{\frac{\pi}{2}}=-\left(\cos \frac{\pi}{2}-\cos 0\right)=-1 m$
We have $v(t)=\cos t, \frac{\pi}{2} \leq t \leq 2 \pi$
Deplacement : $s(t)=\int_{\frac{\pi}{2}}^{2 \pi} \cos t d t=\left.\sin t\right|_{\frac{\tau}{2}} ^{2 \pi}=\left(\sin 2 \pi-\sin \frac{\pi}{2}\right)=-1 m$
Distance :
$$
\begin{aligned}
\int_{\frac{x}{2}}^{2 \pi}|\cos t| d t &=-\int_{\frac{x}{2}}^{\frac{3 x}{2}} \cos t d t+\int_{\frac{3 \pi}{2}}^{0} \cos t d t \\
&=\left.\sin t\right|_{\frac{\pi}{2}} ^{\frac{3 \pi}{2}}+\left.\sin t\right|_{\frac{1 \pi}{2}} ^{2 \pi} \\
&=1-(-2)=[3 m]
\end{aligned}
$$
