Kinematics

JEE Main level, single correct, medium

43 of 308

Question 43 of 308

The relation between time ‘tt’ and distance ‘xx’ is t=αx2+βxt=\alpha {x}^{2}+\beta x, where α\alpha and β\beta are constants. The relation between acceleration (a)\left(a\right) and velocity (v)\left(v\right) is:

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Answer: (A) a=−2αv3a=-2\alpha {v}^{3}

Given: t=αx2+βxt=\alpha {x}^{2}+\beta x
Differentiating w.r.t time, we get
dtdx=2αx+β\dfrac{dt}{dx}=2\alpha x+\beta
⇒1v=2αx+β\Rightarrow \dfrac{1}{v}=2\alpha x+\beta
Differentiating w.r.t time again, we get
⇒−1v2dvdt=2αdxdt\Rightarrow -\dfrac{1}{{v}^{2}}\dfrac{dv}{dt}=2\alpha \dfrac{dx}{dt}
⇒dvdt=−2αv3[asv=dxdt]⇒a=dvdt=−2αv3\Rightarrow \dfrac{dv}{dt}=-2\alpha {v}^{3}\left[\text{as}v=\dfrac{dx}{dt}\right] \Rightarrow a=\dfrac{dv}{dt}=-2\alpha {v}^{3}

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