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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Straight Line

A car starts from rest , moves with an acceleration $a$ and then decelerates at $b$ for some time to come to rest. If the total time taken is t, the maximum velocity is

$\begin{array}{1 1} \large\frac{abt}{a+b} \\ \large\frac{a^2t}{a+b} \\ \large\frac{at}{a+b} \\\large\frac{b^2t}{a+b} \end{array} $

1 Answer

Answer : $ \large\frac{abt}{a+b}$
$v=0+at_1;0=at_1-b(t-t_1)$
or $t_1 =\large\frac{bt}{a+b}$
$\therefore v_{max}=\large\frac{abt}{a+b}$
answered Aug 13, 2014 by meena.p
 

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