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A particle starts with initial speed $u$ and retardation a to come to rest in time $T$. The time taken to cover first half of the total path travelled is .......
$\frac{T}{\sqrt{2}}$
$T\left(1-\frac{1}{\sqrt{2}}\right)$
$\frac{T}{2}$
$\frac{3 T}{4}$
Solution
(b)
Retardation $\rightarrow a$
Initial velocity $\rightarrow u$
$(I)$ For total journey
$v=u+a t$
$0=u-a T$
$\Rightarrow u=a T \ldots (i)$
$d=u T-\frac{1}{2} a T^2$
Dividing by $2$ on both sides
$\frac{d}{2}=\frac{u T}{2}-\frac{1}{2} \frac{a T^2}{2} \ldots (ii)$
On comparing equation $(i)$ and $(iii)$
$\frac{u T}{2}-\frac{1}{2} \frac{a T^2}{2}=u t-\frac{1}{2} a t^2$
Put $u=a T$
$\Rightarrow \frac{a T^2}{2}-\frac{a T^2}{4}=a T t-\frac{1}{2} a t^2$
$\Rightarrow \frac{T^2}{4}=T t-\frac{t^2}{2}$
Multiplying by 4 on both sides
$T^2=4 T t-2 t^2 \Rightarrow 2 t^2-4 T t+T^2=0$
On solving this quadratic equation,
$t=T-\frac{T}{\sqrt{2}} \Rightarrow t=T\left(1-\frac{1}{\sqrt{2}}\right)$
$(II)$ For half journey
$\frac{d}{2}=u t-\frac{1}{2} a t^2 \ldots (iii)$
