A car accelerates from rest at a constant rat | vedclass

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Question

$v _{0}=\alpha t _{1}$ and $0= v _{0}-\beta t _{2} \Rightarrow v _{0}=\beta t _{2}$
$t _{1}+ t _{2}= t$
$v _{0}\left(\frac{1}{\alpha}+\frac{1}{\beta

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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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