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A car accelerates from rest at a constant rate \alpha for sometime after which it decelerates at a constant rate \beta to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
Option: 1 \frac{4\; \alpha \; \beta }{\left ( \alpha +\beta \right ) }\; t^{2}
Option: 2 \frac{2\; \alpha \; \beta }{\left ( \alpha +\beta \right ) }\; t^{2}
Option: 3 \frac{ \alpha \; \beta }{2\left ( \alpha +\beta \right ) }\; t^{2}  
Option: 4 \frac{ \alpha \; \beta }{4\left ( \alpha +\beta \right ) }\; t^{2}

Answers (1)


Let the car accelerate for time t_{1} traveling distance S_{1} and acquire maximum velocity v.
Then,S_{1}=\frac{1}{2} \alpha t_{1}^{2} [from the equation, S=u t+\frac{1}{2} a t^{2} ]
and v=\alpha t_{1} \quad  [from the equation, v=u+a t ]

\Rightarrow \quad S_{1}=\frac{1}{2} \alpha \times \frac{v^{2}}{\alpha^{2}} =\frac{v^{2}}{2 \alpha} ......(1)
After this car decelerates for time t_{2} to come to rest
Hence, \quad S_{2}=v t_{2}-\frac{1}{2} \beta t_{2}^{2}       [from the equation, S=u t-\frac{1}{2} a t^{2} ]
and 0=v-\beta t_{2}         [from the equation, u=v-a t ]
\Rightarrow \quad S_{2}=\frac{v^{2}}{\beta}-\frac{v^{2}}{2 \beta}=\frac{v^{2}}{2 \beta} \ldots (ii)

\begin{array}{lrl}\text { Now, } & t_{1}+t_{2} & =t \\ \\ \Rightarrow & \frac{v}{\alpha}+\frac{v}{\beta}=t \\ \\ \Rightarrow & v=\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\end{array}
Also, total distance traveled,

\begin{aligned} S &=S_{1}+S_{2} \\ &=\frac{v^{2}}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2} \end{aligned}

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Kuldeep Maurya

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