Kinematics

JEE Main level, single correct, medium

16 of 308

Question 16 of 308

Position of an ant ( SS in metres) moving in Y−ZY-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=1st=1s will be :

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Answer: (D) 4ms−14m{s}^{-1} in yy-direction

The velocity of the ant can be found as follows:
v⃗=dS⃗dt=ddt(2t2j^+5k^)=4tj^...(1)\begin{matrix}\vec{v} & = & \dfrac{d\vec{S}}{dt} \\ & = & \dfrac{d}{dt}\left(2{t}^{2}\hat{j}+5\hat{k}\right) \\ & = & 4t\hat{j}...\left(1\right)\end{matrix}
At t=1st=1s, from equation (1), it follows that
v⃗=4j^\vec{v}=4\hat{j}

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