Laws of Motion

JEE Main level, single correct, medium

9 of 210

Question 9 of 210

A 4 kg mass moves under the influence of a force F⃗=(4t3i^−3tj^)N\vec{F}=\left(4 t^{3} \hat{i}-3 t \hat{j}\right) \mathrm{N} where tt is the time in second. If mass starts from origin at t=0t=0, the velocity and position after t=2 st=2 \mathrm{~s} will be:

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Answer: (C) v⃗=4i^−32j^r⃗=85i^−j^\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}-\hat{j}

Given F⃗=(4t3i^−3tj^)\vec{F} = (4t^3\hat{i} - 3t\hat{j}) N, m = 4 kg
a⃗=F⃗m=t3i^−3t4j^\vec{a} = \frac{\vec{F}}{m} = t^3\hat{i} - \frac{3t}{4}\hat{j}
Integrating with v⃗(0)=0\vec{v}(0) = 0:
v⃗=t44i^−3t28j^\vec{v} = \frac{t^4}{4}\hat{i} - \frac{3t^2}{8}\hat{j}
At t=2t = 2: v⃗=164i^−128j^=4i^−32j^\vec{v} = \frac{16}{4}\hat{i} - \frac{12}{8}\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}
Integrating with r⃗(0)=0\vec{r}(0) = 0:
r⃗=t520i^−t38j^\vec{r} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}
At t=2t = 2: r⃗=3220i^−88j^=85i^−j^\vec{r} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}

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