A car accelerates from rest at a constant rate for some time afte… - Vidyadip

MCQ

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

✓

Answer

Correct option: C.

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

c
$v _{0}=\alpha t _{1}$ 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

$=\frac{1}{2} \times t \times v _{0}=\frac{1}{2} \times t \times \frac{\alpha \beta t }{\alpha+\beta}=\frac{\alpha \beta t ^{2}}{2(\alpha+\beta)}$

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