A car accelerates from rest at a constant rate α for some time after which it decelerates at a cons

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2.Motion in Straight Line

hard

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$\frac{4 \alpha \beta}{(\alpha+\beta)} t ^{2}$

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

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

D$\frac{\alpha \beta}{4(\alpha+\beta)} t ^{2}$

(JEE MAIN-2021)

Solution

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

Standard 11

Physics

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hard

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

Solution

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