Laws of Motion

JEE Main level, single correct, medium

18 of 210

Question 18 of 210

At any instant the velocity of a particle of mass 500g500g is (2ti^+3t2j^)ms−1\left(2t\hat{i}+3{t}^{2}\hat{j}\right)m{s}^{-1} . If the force acting on the particle at t=1st=1s is (i^+xj^)N(\hat{i}+x\hat{j})N . Then the value of xx will be:

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Answer: (A) 33

The equation for velocity is
v⃗=(2ti^+3t2j^)\vec{v}=\left(2t\hat{i}+3{t}^{2}\hat{j}\right)
Thus, the acceleration is
a⃗=dv⃗dt=d(2ti^+3t2j^)dt=(2i^+6tj^)\vec{a}=\dfrac{d\vec{v}}{dt}=\dfrac{d(2t\hat{i}+3{t}^{2}\hat{j})}{dt}=\left(2\hat{i}+6t\hat{j}\right)
At t=1st=1s, a⃗=2i^+6j^\vec{a}=2\hat{i}+6\hat{j}
The force is written as F⃗=ma⃗=2i^+6j^2=i^+3j^\vec{F}=m\vec{a}=\dfrac{2\hat{i}+6\hat{j}}{2}=\hat{i}+3\hat{j}
Comparing it with the given value of force, x=3x=3.

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