Kinematics

JEE Main level, numerical answer, medium

55 of 308

Question 55 of 308

A body of mass 2kg2kg moves under a force of (2i^+3j^+5k^)N\left(2\hat{i}+3\hat{j}+5\hat{k}\right)N It starts from rest and was at the origin initially. After 4s4s, its new coordinates are (8,b,20)\left(8,b,20\right). The value of bb is ______.
(Round off to the Nearest Integer)

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Answer: 12

a⃗=F⃗m=2i^+3j^+5k^2\vec{a}=\dfrac{\vec{F}}{m}=\dfrac{2\hat{i}+3\hat{j}+5\hat{k}}{2}
=i^+1.5j^+2.5k^=\hat{i}+1.5\hat{j}+2.5\hat{k}
τ⃗=u⃗t+12a⃗t2\vec{\tau }=\vec{u}t+\dfrac{1}{2}\vec{a}{t}^{2}
=0+12(i^+1.5j^+2.5k^)(16)=0+\dfrac{1}{2}\left(\hat{i}+1.5\hat{j}+2.5\hat{k}\right)\left(16\right)
=8i^+12j^+20k^=8\hat{i}+12\hat{j}+20\hat{k}
b=12b=12

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