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A car staring from rest acceleration at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at rate $\large\frac{f}{2}$ to come to rest. If the total distance covered is $15\;s$ , then

$\begin{array}{1 1} s= \large\frac{ft^2}{72} \\ s= \large\frac{ft^2}{4} \\ s= \large\frac{ft^2}{6} \\s= \large\frac{ft^2}{2} \end{array} $

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Answer : $ s= \large\frac{ft^2}{72}$
$s= v_0f_1$ and $v_0 2t_1 =2s$
Distance moved with uniform speed $(15-3)s= 12s$
$v_o = \sqrt {2sf} :12s= v_o t$
$12s = t \sqrt {2sf}$
or $ s= \large\frac{ft^2}{72}$
answered Aug 13, 2014 by meena.p

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