A car, starting from rest, accelerates at the rate $\mathrm{f}$ through a distance $\mathrm{S}$, then…

A car, starting from rest, accelerates at the rate $\mathrm{f}$ through a distance $\mathrm{S}$, then continues at constant speed for time $\mathrm{t}$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15 S$, then
  1. $S=\frac{1}{6} f t^{2}$
  2. $S=f t$
  3. $S=\frac{1}{4} f t^{2}$
  4. $S=\frac{1}{72} f t^{2}$

Solution

Distance from $A$ to $B=S=\frac{1}{2} f t_{1}^{2}$ Distance from $B$ to $C=\left(f t_{1}\right) t$
Distance from $C$ to $D=\frac{u^{2}}{2 a}=\frac{\left(f t_{1}\right)^{2}}{2(f / 2)}=f t_{1}^{2}=2 S$


$\Rightarrow \quad S+f t_{1} t+2 S=15 S$
$\Rightarrow \quad f t_{1} t=12 S$
............. (i)
$\frac{1}{2} f t_{1}^{2}=S$
Dividing (i) by (ii), we get $t_{1}=\frac{\mathrm{t}}{6}$ $\Rightarrow S=\frac{1}{2} f\left(\frac{t}{6}\right)^{2}=\frac{f t^{2}}{72}$

Asked in: JEE Mains - Motion In One Dimension - Test 2

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