Question

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

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

Answer 1

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

Initial velocity of the particle is $\left(4\hat{j}\right)m/s$
at $time=1s$ acceleration of the particle is $6m/s^2$ along x-axis
Acceleration of the particle is constant

$\overrightarrow{x}=3t^2\overrightarrow{i}+4t\overrightarrow{j}+\frac{1}{2}\overrightarrow{k}\overrightarrow{v}=\frac{dx}{dt}=6t\overrightarrow{i}+4\overrightarrow{j}\overrightarrow{a}=\frac{dv}{dt}=6\overrightarrow{i}$

Clearly acceleration is constant, $6m/s^2$ as it is not dependent on any variable.
Put $t=0$ in the above equations to get initial values of velocity i.e. $4\overrightarrow{j}$m/s