A body with mass 5 kg is acted upon by a force F = (–3î +4 ĵ)N . If its initial velocity at t = 0 is v =( 6 î -12 ĵ ) ms⁻¹ , the time at which it will just have a velocity along the y-axis is

The correct answer is 10 s.

To find the time at which the body has velocity only along the y-axis, we need to analyze its motion in both the x and y directions. Let's list the given parameters:
Mass of the body, $m=5\text{kg}$
Force acting on the body, $\overrightarrow{F}=-3\hat{i}+4\hat{j}\text{N}$
Initial velocity at time $t=0$, $\overrightarrow{u}=6\hat{i}-12\hat{j}{\text{ms}}^{-1}$

First, we find the acceleration vector $\overrightarrow{a}$ using Newton's second law:
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}$
$\overrightarrow{a}=\frac{-3\hat{i}+4\hat{j}}{5}=-0.6\hat{i}+0.8\hat{j}{\text{ms}}^{-2}$

Using the first equation of motion in vector form, the velocity vector $\overrightarrow{v}$ at any time $t$ is:
$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
$\overrightarrow{v}=\left(6\hat{i}-12\hat{j}\right)+\left(-0.6\hat{i}+0.8\hat{j}\right)t$
Grouping the x-axis ($\hat{i}$) and y-axis ($\hat{j}$) components:
$\overrightarrow{v}=\left(6-0.6t\right)\hat{i}+\left(-12+0.8t\right)\hat{j}$

For the body to have a velocity pointing strictly along the y-axis, the horizontal component of the velocity (along the x-axis) must be equal to zero:
$v_x=6-0.6t=0$
$0.6t=6$
$t=\frac{6}{0.6}=10\text{s}$

Thus, the time at which the velocity of the body will be entirely along the y-axis is 10 s.