A car accelerates from rest at a constant rat | vedclass

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(C) Let $t _{1}$ be the time of acceleration and $t _{2}$ be the time of deceleration. The maximum velocity reached is $v _{0}$.From the equations of

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$\frac{\alpha \beta}{2(\alpha+\beta)} t ^{2}$

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(C) Let $t _{1}$ be the time of acceleration and $t _{2}$ be the time of deceleration. The maximum velocity reached is $v _{0}$.From the equations of motion:$v _{0} = \alpha t _{1} \Rightarrow t _{1} = \frac{v _{0}}{\alpha}$$0 = v _{0} - \beta t _{2} \Rightarrow t _{2} = \frac{v _{0}}{\beta}$Given that the total time is $t = t _{1} + t _{2}$,we have:$t = \frac{v _{0}}{\alpha} + \frac{v _{0}}{\beta} = v _{0} \left( \frac{\alpha + \beta}{\alpha \beta} ight)$Thus,the maximum velocity is $v _{0} = \frac{\alpha \beta t}{\alpha + \beta}$.The total distance travelled is the area under the $v-t$ graph,which is a triangle with base $t$ and height $v _{0}$:Distance $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes t 	imes v _{0}$Substituting $v _{0}$:Distance $= \frac{1}{2} 	imes t 	imes \left( \frac{\alpha \beta t}{\alpha + \beta} ight) = \frac{\alpha \beta t ^{2}}{2(\alpha + \beta)}$

$A$ 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

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