Question

A car starts from a point $P$ at time $t=0\text{sec}$ and stops at point $Q$. The distance $S$, in meters,travelled by it in $t\text{sec}$ is given by $S=t^2(2-\frac{t}{3}).$ Which of the following(s) is/are true

• A. Time taken to reach $Q$ is $4\text{sec}$
• B. Distance between $P$ and $Q$ is $32\text{meters}$
• C. Acceleration at $t=1\text{sec},$ is $2\text{m/}{\text{sec}}^2$
• D. Velocity at $t=1\text{sec},$ is $3\text{m/sec}$

Answer 1

Answer: (A), (C), (D)

Velocity at $t=1\text{sec},$ is $3\text{m/sec}$

Given : $S=t^2(2-\frac{t}{3})=2t^2-\frac{t^3}{3}$
$\Rightarrow \text{velocity},V=\frac{dS}{dt}=4t-t^2\cdots \left(1\right)$
$\Rightarrow \text{acceleration},a=\frac{dV}{dt}=4-2t\cdots \left(2\right)$
$\left(a\right)$ At point $Q,V=0\Rightarrow t(4-t)=0\Rightarrow t=0,4$
So, car reaches at $Q$ in $t=4\text{sec}$
$\left(b\right)$ Distance between $P$ and $Q=32-\frac{64}{3}=\frac{32}{3}\text{meters}$
$\left(c\right)$ Acceleration,$a(t=1)=4-2=2\text{m/}{\text{sec}}^2$
$\left(d\right)$ Velocity,$V(t=1)=4-1=3\text{m/sec}$