Question

The position of a car is given by the equation $f\left(t\right)=\sin \left(\pi t\right)+3t^2-10t-18$. Find the car's acceleration when $t=4$.

Answer 1

We have,

$f\left(t\right)=\sin \left(\pi t\right)+3t^2-10t-18$

On differentiating w.r.t $t$, we get

$v=f^{\prime }\left(t\right)=\pi \cos \left(\pi t\right)+6t-10-0$

$v=f^{\prime }\left(t\right)=\pi \cos \left(\pi t\right)+6t-10$

Again, differentiating w.r.t $t$, we get

$a=f^{\prime \prime }\left(t\right)=-{\pi }^2\sin \left(\pi t\right)+6-0$

$a=f^{\prime \prime }\left(t\right)=-{\pi }^2\sin \left(\pi t\right)+6$

Since, $t=4$

Therefore, the acceleration will be

$=-{\pi }^2\sin \left(4\pi \right)+6$

$=0+6$

$=6$

Hence, this is the answer.