Answer
$\text {(a) velocity = speed = 0.5 m/s}$
$\text {acceleration = $-\frac{\sqrt{3}\pi}{6}$ m/$s^2$ }$
$\text {position = $\frac{3}{\pi}(\frac{\sqrt{3}}{2}-1)$ m}$
$\text {(b) velocity = speed = 1.75 m/s}$
$\text {acceleration = 4 m/$s^2$ }$
$\text {position = 1.283 m}$
Work Step by Step
$\text {It is given that}$
$\text {(a)}$
\begin{align}
v(t) = \cos {\frac{\pi}{3}t}; s = 0m \ when \ t = 1.5 s
\end{align}
$\text {Velocity at t = 1s:}$
\begin{align}
v(1) = \cos {\frac{\pi}{3}} = \frac{1}{2} m/s
\end{align}
$\text {Speed at t = 1s:}$
\begin{align}
v(1) = | \cos {\frac{\pi}{3}}| = \frac{1}{2} m/s
\end{align}
$\text {Acceleration at t = 1s:}$
\begin{align}
& a(t) = v'(t) = -\frac{\pi}{3} \times \sin {\frac{\pi}{3}t} \\
& a(1) = -\frac{\pi}{3} \times \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}\pi}{6}
\end{align}
$\text {Position at t = 1s:}$
\begin{align}
s(t) = \int v(t) \ dt = \int \cos {\frac{\pi}{3}t} \ dt = \frac{3}{\pi}\sin {\frac{\pi}{3}t} + C
\end{align}
$\text{By appying the initially given conditions, we can find C:}$
\begin{align}
& 0 = \frac{3}{\pi}\sin {\frac{\pi}{2}} + C \\
& 0 = \frac{3}{\pi} + C \\
& C = -\frac{3}{\pi} \\
s(t) = &\frac{3}{\pi}\sin {\frac{\pi}{3}t} -\frac{3}{\pi} \\
s(1) = &\frac{3}{\pi}\sin {\frac{\pi}{3}} -\frac{3}{\pi} = \frac{3}{\pi}(\frac{\sqrt{3}}{2}-1)
\end{align}
$\text {It is given that}$
$\text {(b)}$
\begin{align}
a(t) = 5t - t^3; s = 1m \ \ and \ \ v = -0.5 m/s \ \ when \ \ t = 0s
\end{align}
$\text {Acceleration at t = 1s:}$
\begin{align}
a(1) = 5-1=4m/s^2
\end{align}
$\text {Velocity at t = 1s:}$
\begin{align}
v(t) = \int a(t) \ dt = \int (5t - t^3) \ dt = \frac{5t^2}{2}-\frac{t^4}{4} + C_1
\end{align}
$\text{By appying the initially given conditions, we can find $C_1$:}$
\begin{align}
&C_1 = -0.5 \\
v(t) = &\frac{5t^2}{2}-\frac{t^4}{4} - 0.5 \\
&v(1) = 1.75 m/s
\end{align}
$\text {Speed at t = 1s:}$
\begin{align}
v(1) = |\frac{5t^2}{2}-\frac{t^4}{4} - 0.5| = 1.75m/s
\end{align}
$\text {Position at t = 1s:}$
\begin{align}
s(t) = \int v(t) \ dt = \int (\frac{5t^2}{2}-\frac{t^4}{4} - 0.5) \ dt = \frac{5t^3}{6}-\frac{t^5}{20} - 0.5t + C_2
\end{align}
$\text{By appying the initially given conditions, we can find $C_2$:}$
\begin{align}
& C_2 = 1 \\
s(t) = &\frac{5t^3}{6}-\frac{t^5}{20} - 0.5t + 1 \\
& s(1) = 1.283m
\end{align}
