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

(D) $4 \mathrm{~m} / \mathrm{s}$ in $y$-direction

(A) $16 \mathrm{~m} / \mathrm{s}$ in $y$-direction

(B) $4 \mathrm{~m} / \mathrm{s}$ in x-direction

(C) $9 \mathrm{~m} / \mathrm{s}$ in z-direction

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Kinematics

Position of an ant (S in metres) moving in Y-Z plane is given by

S = 2 t^{2} \hat{j} + 5 \hat{k}

(where t is in second). The magnitude and direction of velocity of the ant at

t = 1 s

will be :

(A) $16 m / s$ in $y$ -direction
(B) $4 m / s$ in x-direction
(C) $9 m / s$ in z-direction
(D) $4 m / s$ in $y$ -direction✓ Correct

Answer: (D)

Solution

The velocity is the time derivative of the position vector:

v = \frac{d s}{d t} = \frac{d}{d t} (2 t^{2} \hat{j} + 5 \hat{k}) = 4 t \hat{j}

t = 1

v = 4 \hat{j}

m/s, so the speed is

4

m/s directed along the

y

-axis.

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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