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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A car acceleration from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If t is the total time elapsed , then the maximum velocity attained by the car is

$\begin{array}{1 1} \alpha \beta t \\ \large\frac{t}{\alpha \beta } \\ \large\frac{\alpha \beta t}{\alpha+ beta} \\\large\frac{\alpha+ \beta t}{\alpha \beta}\end{array} $

Can you answer this question?
 
 

1 Answer

+1 vote
Answer : $ \large\frac{\alpha \beta t}{\alpha+ \beta}$
$V=0+\alpha t_1$
$t_1=\large\frac{V}{\alpha}+\frac{V}{\beta}$
Similarity, $t_2=\large\frac{V}{\beta}$
or $t_1+t_2= \large\frac{V}{\alpha} +\frac{V}{\beta}$
$\qquad= \large\frac{\alpha \beta t}{\alpha+ \beta}$
answered Aug 19, 2014 by meena.p
 

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