Question

Position vector of a particle at time = t sec is given by $\left(3t^2\right)\hat{i}+\left(4t\right)\hat{j}+\left(\frac{1}{2}\right)\hat{k}m$. Choose the correct option(s).

• A. Initially the particle is in the $xy$ plane
• B. Initial velocity of the particle is $\left(4\hat{j}\right)m/s$
• C. At time $t=1s$ acceleration of the particle is $6m/s^2$ along x axis
• D. Acceleration of the particle is constant

Answer 1

Answer: (B), (C), (D)

The correct options are
B At time $t=1s$ acceleration of the particle is $6m/s^2$ along x axis
C Acceleration of the particle is constant
D Initial velocity of the particle is $\left(4\hat{j}\right)m/s$

A: Because of k component, velocity is in z-direction also.
B: Differentiate to get $\overrightarrow{v}=6t\hat{i}+4\hat{j}$

Thus at $t=0,v=4\hat{j}$
C: Double differentiate to get $\overrightarrow{a}=6\hat{i}$

Thus at $t=1,a=6\hat{i}$
D: As found in part C, acceleration is independent of time.