Question

The position vector of a particle is given by $\vec{x}=(t^{3}-3t^{2}+2)\hat{i}$, the time at which the velocity of the moving particle becomes zero is

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s
• E. 5 s

Hint:

"Velocity is zero" physically means the particle is momentarily at rest, which often occurs at the turning point of motion.

Answer 1

The Correct Option is B

Concept:
Physics - Kinematics and Derivatives.
Velocity is the first derivative of position with respect to time: $v = \frac{dx}{dt}$.
Step 1: Differentiate the position function.
Given position $x = t^3 - 3t^2 + 2$. $$ v = \frac{d}{dt}(t^3 - 3t^2 + 2) $$ $$ v = 3t^2 - 6t $$
Step 2: Set the velocity to zero.
The problem asks for the time when velocity becomes zero: $$ 3t^2 - 6t = 0 $$
Step 3: Solve the quadratic equation.
Factor out $3t$: $$ 3t(t - 2) = 0 $$ This gives two solutions: $$ t = 0 \text{ s} \quad \text{or} \quad t = 2 \text{ s} $$
Step 4: Select the relevant time.
Since $t=0$ is the start of motion and $t=2$ is an option provided, the particle's velocity becomes zero at $t = 2$ s.