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Physics · Motion in one Dimension

JEE Main 2024 — 27 January, Shift 1 — Question 29

Position of an ant ( S in metres) moving in Y−Z\mathrm{Y}-\mathrm{Z} plane is given by S=2t2j^+5k^S=2 t^{2} \hat{j}+5 \hat{k} (where tt is in second). The magnitude and direction of velocity of the ant at t=1 st=1 \mathrm{~s} will be :

  1. Option A:

    16 m/s16 \mathrm{~m} / \mathrm{s} in y-direction

  2. Option B:

    4 m/s4 \mathrm{~m} / \mathrm{s} in x -direction

  3. Option C:

    9 m/s9 \mathrm{~m} / \mathrm{s} in z -direction

  4. Option D:

    4 m/s4 \mathrm{~m} / \mathrm{s} in y-direction

    Correct

Answer: D

Step-by-step solution

v⃗=ds⃗dt=4tj^\vec{v}=\frac{d \vec{s}}{d t}=4 t \hat{j}

At t=1sec⁡v→=4j^\mathrm{t}=1 \sec \overrightarrow{\mathrm{v}}=4 \hat{\mathrm{j}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Physics
Chapter
Motion in one Dimension
Topic
Uniformly Accelerated Motion

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