Answer
Distance: $-1+\frac{\sqrt{2}}{2}+\frac{\pi}{4} \mathrm{m}$ or about $0.4925 \mathrm{m}$
Displacement: $-1+\frac{\sqrt{2}}{2}+\frac{\pi}{4} \mathrm{m}$ or about $0.4925 \mathrm{m}$
Work Step by Step
Given:
$$
\begin{array}{c}
\sin (t)=a(t) \\
1=v_{0} \\
\frac{\pi}{4} \leq t \leq \frac{\pi}{2}
\end{array}
$$
The velocity is the integral of the acceleration.
$$
v(t)=\int a(t) d t+v_{0}=\int 1+\sin (t) d t=1-\cos t
$$
The displacement is the integral of the velocity:
$$
\begin{aligned}
s(t)=\int_{\pi / 4}^{\pi / 2} v(t) d t &=\left.(-\sin (t)+t)\right|_{\pi / 4} ^{\pi / 2} \\
=-\sin \frac{\pi}{2}+\frac{\pi}{2}+\sin \frac{\pi}{4}-\frac{\pi}{4} \\
=\frac{\sqrt{2}}{2}+\frac{\pi}{4}-1 \\
\approx 0.4925 &
\end{aligned}
$$
The distance is the integral of the velocity:
$$
\begin{aligned}
s(t)=\int_{\pi / 4}^{\pi / 2} v(t) d t &=\left.(-\sin (t)+t)\right|_{\pi / 4} ^{\pi / 2} \\
=\frac{\pi}{2}+\sin \frac{\pi}{4}-\frac{\pi}{4}-\sin \frac{\pi}{2} \\
=\frac{\sqrt{2}}{2}+\frac{\pi}{4}-1 \\
\approx 0.4925 &
\end{aligned}
$$
