Kinematics

JEE Main level, single correct, medium

28 of 308

Question 28 of 308

Two buses PP and QQ start from a point at the same time and move in a straight line and their positions are represented by xP(t)=αt+βt2{x}_{P}\left(t\right)=\alpha t+\beta {t}^{2} and xQ(t)=ft−t2{x}_{Q}\left(t\right)=ft-{t}^{2}. At what time, both the buses have same velocity ?

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Answer: (D) f−α2(1+β)\dfrac{f-\alpha }{2\left(1+\beta \right)}

As we know, v=dxdtv=\dfrac{dx}{dt}.
Given, xP(t)=αt+βt2{x}_{P}\left(t\right)=\alpha t+\beta {t}^{2}
Velocity of bus PP is vP=dxPdt=α+2βt{v}_{P}=\dfrac{d{x}_{P}}{dt}=\alpha +2\beta t and velocity of bus QQ is vQ=f−2t{v}_{Q}=f-2t
When vP=vQ{v}_{P}={v}_{Q}
⇒α+2βt=f−2t\Rightarrow \alpha +2\beta t=f-2t
⇒f−α=2(β+1)t⇒t=f−α2(β+1)\Rightarrow f-\alpha =2\left(\beta +1\right)t \Rightarrow t=\dfrac{f-\alpha }{2\left(\beta +1\right)}

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