Laws of Motion

JEE Main level, single correct, medium

15 of 210

Question 15 of 210

The position vector of a particle related to time tt is given by r⃗=(10ti^+15t2j^+7k^)m\vec{r}=\left(10t\hat{i}+15{t}^{2}\hat{j}+7\hat{k}\right)m. The direction of net force experienced by the particle is :

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Answer: (C) Positive yy-axis

Given, the position vector of the particle is
r⃗=10ti^+15t2j^+7k^\vec{r}=10t\hat{i}+15{t}^{2}\hat{j}+7\hat{k}
The velocity of the particle can be calculated as follows:
v⃗=dr⃗dt=ddt(10ti^+15t2j^+7k^)=10i^+30tj^\begin{matrix}\vec{v} & = & \dfrac{d\vec{r}}{dt} \\ & = & \dfrac{d}{dt}\left(10t\hat{i}+15{t}^{2}\hat{j}+7\hat{k}\right) \\ & = & 10\hat{i}+30t\hat{j}\end{matrix}
And, the acceleration of the particle is given by
a⃗=dv⃗dt=ddt(10i^+30tj^)=30j^\begin{matrix}\vec{a} & = & \dfrac{d\vec{v}}{dt} \\ & = & \dfrac{d}{dt}\left(10\hat{i}+30t\hat{j}\right) \\ & = & 30\hat{j}\end{matrix}
Hence, the force on the particle can be written as
F⃗=ma⃗=30mj^\begin{matrix}\vec{F} & = & m\vec{a} \\ & = & 30m\hat{j}\end{matrix}

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