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Physics · Motion in one Dimension

JEE Main 2026 — 21 January, Morning Shift — Question 30

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) N where tt is the time in second. If mass starts from origin at t=0\mathrm{t}=0, the velocity and position after t=2 s\mathrm{t}=2 \mathrm{~s} will be :

  1. Option A:

    v⃗=3i^+32j^r⃗=65i^+j^\vec{v}=3 \hat{i}+\frac{3}{2} \hat{j} \vec{r}=\frac{6}{5} \hat{i}+\hat{j}

  2. Option B:

    v⃗=4i^−32j^r⃗=85i^−j^\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \vec{r}=\frac{8}{5} \hat{i}-\hat{j}

    Correct
  3. Option C:

    v⃗=4i^+52j^r⃗=85i^+2j^\vec{v}=4 \hat{i}+\frac{5}{2} \hat{j} \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}

  4. Option D:

    v⃗=4i^−32j^r⃗=65i^−j^\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \vec{r}=\frac{6}{5} \hat{i}-\hat{j}

Answer: B

Step-by-step solution

F⃗=4t3i^−3tj^\vec{F}=4 t^{3} \hat{i}-3 t \hat{j}

a⃗=F⃗m=t3i^−34j^ax=t3dvxdt=t3∫vx=0vx2dvx=∫t=0t=2t3dtvx2−0=[t44]02vx2=4\begin{aligned} & \vec{a}=\frac{\vec{F}}{m}=t^{3} \hat{i}-\frac{3}{4} \hat{j} & a_{x}=t^{3} & \frac{d v_{x}}{d t}=t^{3} & \int_{v_{x}=0}^{v_{x_{2}}} d v_{x}=\int_{t=0}^{t=2} t^{3} d t & v_{x_{2}}-0=\left[\frac{t^{4}}{4}\right]_{0}^{2} & v_{x_{2}}=4 \end{aligned} ay=−34tdvydt=−34∫0vy2dvy=∫02−34tdtvy2=−34[t22]02vy2=−3232j^\begin{aligned} & a_{y}=\frac{-3}{4} t & \frac{d v_{y}}{d t}=-\frac{3}{4} & \int_{0}^{v_{y_{2}}} d v_{y}=\int_{0}^{2} \frac{-3}{4} t d t & v_{y_{2}}=\frac{-3}{4}\left[\frac{t^{2}}{2}\right]_{0}^{2} & v_{y_{2}}=\frac{-3}{2} & \frac{3}{2} \hat{j} \end{aligned} vx=t44∫0x2dx=∫02t44dtx2−0=[t520]02x2=85\begin{aligned} & \mathrm{v}_{\mathrm{x}}=\frac{\mathrm{t}^{4}}{4} & \int_{0}^{\mathrm{x}_{2}} \mathrm{dx}=\int_{0}^{2} \frac{\mathrm{t}^{4}}{4} \mathrm{dt} & \mathrm{x}_{2}-0=\left[\frac{\mathrm{t}^{5}}{20}\right]_{0}^{2} & \mathrm{x}_{2}=\frac{8}{5} \end{aligned} vy=−38t2∫0y2dx=−38t2dty2−0=−38[t33]02y2=−1\begin{aligned} & v_{y}=\frac{-3}{8} t^{2} & \int_{0}^{y_{2}} d x=\frac{-3}{8} t^{2} d t & y_{2}-0=\frac{-3}{8}\left[\frac{t^{3}}{3}\right]_{0}^{2} & y_{2}=-1 \end{aligned} r⃗=85i^−j^\vec{r}=\frac{8}{5} \hat{i}-\hat{j}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Motion in one Dimension
Topic
Uniformly Accelerated Motion