Answer
(a)
Acceleration at time $t=5$ min is $-120\hspace{4px} m/s^{2}$.
(b)
The acceleration in the graph is negative. So, the helicopter is slowing down during this time interval.
Work Step by Step
(a)
Firstly, calculate first derivative of $s(t) = 300t − 4t^{3}$ m using power rule.
$\dfrac{ds}{dt}=300-4\times3t^{2}$
Now calculate second derivative by differentiating on both side of equation $\dfrac{ds}{dt}=300-4\times3t^{2}$.
$\dfrac{d^{2}s}{dt^{2}}=0-4\times3\times2t=-24t$
Now, substitute $t = 5$ min and solve.
$\dfrac{d^{2}s}{dt^{2}}=-24\times 5=-120$
Since acceleration is the second derivative of displacement.
Acceleration at time $t=5$ min is $-120\hspace{4px} m/s^{2}$.
(b)
The acceleration in the graph is negative.
Thus, the velocity must be decreasing.
So, the helicopter is slowing down during this time interval.
