Question 98 of 308

A car accelerates from rest at a constant rate $\alpha$ for some time 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:

A$\dfrac{4\alpha \beta }{(\alpha +\beta )}{t}^{2}$B$\dfrac{2\alpha \beta }{(\alpha +\beta )}{t}^{2}$C$\dfrac{\alpha \beta }{2(\alpha +\beta )}{t}^{2}$D$\dfrac{\alpha \beta }{4(\alpha +\beta )}{t}^{2}$

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Answer: (C) $\dfrac{\alpha \beta }{2(\alpha +\beta )}{t}^{2}$

${v}_{0}=\alpha {t}_{1}$ and $0={v}_{0}-\beta {t}_{2}\Rightarrow {v}_{0}=\beta {t}_{2}$
${t}_{1}+{t}_{2}=t$
${v}_{0}\left(\dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)=t$
$\Rightarrow {v}_{0}=\dfrac{\alpha \beta t}{\alpha +\beta }$
Distance $=$ area of $v-t$ graph
$=\dfrac{1}{2}\times t\times {v}_{0}=\dfrac{1}{2}\times t\times \dfrac{\alpha \beta t}{\alpha +\beta }=\dfrac{\alpha \beta {t}^{2}}{2(\alpha +\beta )}$