A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$, then
A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$, then
$S = \frac{1}{2}f{t^2}$
$S = \frac{1}{4}f{t^2}$
$S = \frac{1}{{72}}f{t^2}$
$S = \frac{1}{6}f{t^2}$
A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$, then
Let car starts from point $A$ from rest and moves up to point $B$ with acceleration $f$
Velocity of car at point $B$, $v = \sqrt {2fS} $
$As \,{v^2} = {u^2} + 2as]$
Car moves distance $BC$ with this constant velocity in time $t$
$x = \sqrt {2fS} \,.\,t$ ......(i) [As $s = ut$]
So the velocity of car at point $C$ also will be $\sqrt {2fs} $ and finally car stops after covering distance y.
Distance $CD ⇒y = \frac{{{{(\sqrt {2fS} )}^2}}}{{2(f/2)}}$$ = \frac{{2fS}}{f} = 2S$....(ii) $[{\rm{As }}{v^{\rm{2}}} = {u^2} - 2as\, \Rightarrow \,s = {u^2}/2a]$
So, the total distance $AD$ = $AB + BC + CD=15S$ (given)
$⇒$ $S + x + 2S = 15S$ $⇒ x = 12S$
Substituting the value of x in equation (i) we get
$x = \sqrt {2fS} \,.\,t ⇒ 12S = \sqrt {2fS} .t ⇒ 144{S^2} = 2fS.{t^2}$
$⇒$ $S = \frac{1}{{72}}f{t^2}$.
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