Kinematics

JEE Main level, numerical answer, medium

15 of 308

Question 15 of 308

A particle starts from origin at t=0t=0 with a velocity 5i^ms−15\hat{i}m{s}^{-1} and moves in x−yx-y plane under action of a force which produces a constant acceleration of (3i^+2j^)ms−2(3\hat{i}+2\hat{j})m{s}^{-2}. If the xx-coordinate of the particle at that instant is 84m84m, then the speed of the particle at this time is αms−1\sqrt{\alpha }m{s}^{-1}. The value of α\alpha is _______.

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Answer: 673

Given, ux=5ms−1,ax=3ms−2,x=84m{u}_{x}=5m{s}^{-1},{a}_{x}=3m{s}^{-2},x=84m
The formula to calculate the velocity of the particle along x-axis is given by
vx2=ux2+2ax...(1){v}_{x}^{2}={u}_{x}^{2}+2ax...\left(1\right)
From equation (1), it follows that
vx2=25+2(3)(84)⇒vx=23ms−1{v}_{x}^{2}=25+2(3)(84) \Rightarrow {v}_{x}=23m{s}^{-1}
Also, the velocity of the particle along x-axis can be written as
vx=ux+axt...(2){v}_{x}={u}_{x}+{a}_{x}t...\left(2\right)
From equation (2), it follows that
t=23−53=6s\begin{matrix}t & = & \dfrac{23-5}{3} \\ & = & 6s\end{matrix}
Similarly, for the y−y- component of the velocity, it follows that
vy=0+ayt=0+2×(6)=12ms−1\begin{matrix}{v}_{y} & = & 0+{a}_{y}t \\ & = & 0+2\times (6) \\ & = & 12m{s}^{-1}\end{matrix}
Hence, the magnitude of the velocity is given by
v2=vx2+vy2=232+122=673v=673ms−1{v}^{2}={v}_{x}^{2}+{v}_{y}^{2} ={23}^{2}+{12}^{2} =673 v=\sqrt{673}m{s}^{-1}
Therefore, α=673\alpha =673.

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