Question

A particle is moving along $x$ – axis whose position is given by $x=4-9t+\frac{t^3}{3}$ . Mark the correct statement(s) in relation to its motion.

• A. Direction of motion is not changing at any of the instants
• B. Direction of motion is changing at $t=3\text{s}$
• C. For $0<t<3\text{s}$, the particle is slowing down
• D. For $0<t<3\text{s}$ the particle is speeding up

Answer 1

Answer: (B), (C)

For $0<t<3\text{s}$, the particle is slowing down

$x=4-9t+\frac{t^3}{3}$
Differentiating w.r.t $t$, we get
$v=0-9+t^2$
$\Rightarrow v=t^2-9$
Putting $v=0$, we get
$t=3\text{s}$
$\therefore$ Direction of motion is changing at $t=3\text{s}$
Differentiating $v$ w.r.t $t$, we get
$a=2t$
We know that, if $v$ and $a$ are in opposite direction then the particle slows down.
For $0<t<3\text{s}$, $v$ is in negative direction whereas $a$ is positive , so the particle is slowing down.