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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2026 · 21 January · Morning Shift · Physics · Laws of Motion

A 4 kg mass moves under the influence of a force F =(4 t^ 3 i -3 t...

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:

Options

  1. Av⃗=4i^−32j^r⃗=65i^−j^\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}-\hat{j}
  2. Bv⃗=3i^+32j^r⃗=65i^+j^\vec{v}=3 \hat{i}+\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}+\hat{j}
  3. Cv⃗=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}
  4. Dv⃗=4i^+52j^r⃗=85i^+2j^\vec{v}=4 \hat{i}+\frac{5}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}

Correct 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}

Step-by-step solution

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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