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(D) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (-3 \hat{i} + 4 \hat{j}) \ N$,Initial velocity $\vec{u} = (6 \hat{i} - 12 \hat{j}) \ ms^{-1}$.Using Newt

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(D) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (-3 \hat{i} + 4 \hat{j}) \ N$,Initial velocity $\vec{u} = (6 \hat{i} - 12 \hat{j}) \ ms^{-1}$.Using Newton's second law,$\vec{a} = \frac{\vec{F}}{m} = \left(\frac{-3}{5} \hat{i} + \frac{4}{5} \hat{j}ight) \ ms^{-2}$.The velocity at any time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$.$\vec{v} = (6 \hat{i} - 12 \hat{j}) + \left(-\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}ight)t$.$\vec{v} = \left(6 - \frac{3}{5}tight) \hat{i} + \left(-12 + \frac{4}{5}tight) \hat{j}$.For the velocity to be along the $y$-axis,the $x$-component of the velocity must be zero:$6 - \frac{3}{5}t = 0$.$\frac{3}{5}t = 6$.$t = \frac{6 	imes 5}{3} = 10 \ s$.

An object with mass $5 \ kg$ is acted upon by a force,$\vec{F}=(-3 \hat{i}+4 \hat{j}) \ N$. If its initial velocity at $t=0$ is $\vec{v}=(6 \hat{i}-12 \hat{j}) \ ms^{-1}$,the time at which it will just have a velocity along the $y$-axis is: (in $s$)

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