Question Type 1: Deriving to find the rate of change of a kinematic variable Exercises

A particle is thrown vertically upward with initial velocity $v(0)=20\,\mathrm{m/s}$ under constant acceleration $a=-9.8\,\mathrm{m/s^2}$. Given $s(0)=0$, find the time when it returns to $s=0$.

The acceleration of a particle is $a(t)=6t-4$. Given $v(0)=3$ and $s(0)=2$, find expressions for $v(t)$ and $s(t)$.

Given the displacement function

For $s(t)=t e^{-t}$, find expressions for $v(t)$ and $a(t)$.

Given $v(t)=5e^{-2t}$ and $s(0)=0$, find the displacement $s(t)$.

The velocity is $v(t)=\cos(2t)-2\sin t$ and $s(0)=1$. Find $s(t)$.

Given the velocity $v(t)=t^3-3t^2+t+10,$ find the total distance travelled between $t=0$ and $t=5$.

For $s(t)=4t-t^2+3e^t$, find the time when the acceleration $a(t)$ is zero.

Given $s(t)=3\sin t -t^2$, find $v(t)$, $a(t)$, and the time when $a(t)=0$ for $t>0$.

For the displacement $s(t)=e^{-t}+\frac{\sin(-t)}{t}$, find the first time $t>0$ when the object is at rest.

An object moves with displacement

The acceleration is $a(t)=12t^2-6t+1$ and at $t=1$ the velocity is $v(1)=2$, displacement $s(1)=5$. Find $v(t)$ and $s(t)$.