Question

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

• A. $\frac{T}{\sqrt{2}}$
• B. $T\left(1-\frac{1}{\sqrt{2}}\right)$
• C. $\frac{T}{2}$
• D. $\frac{3T}{4}$

Answer 1

Answer: (B)

Retardation $\to a$
Initial velocity $\to u$
(I) For total journey
$v=u+at$
$0=u-aT$
$\Rightarrow u=aT$
$d=uT-\frac{1}{2}a{T}^2$
Dividing by 2 on both sides
$\frac{d}{2}=\frac{uT}{2}-\frac{1}{2}\frac{a{T}^2}{2}$ ...(ii)
(II) For half journey
$\frac{d}{2}=ut-\frac{1}{2}a{t}^2$ ...(iii)
On comparing equation (i) $\&$ (iii)
$\frac{uT}{2}-\frac{1}{2}\frac{a{T}^2}{2}=ut-\frac{1}{2}a{t}^2$
Put $u=aT$
$\Rightarrow \frac{a{T}^2}{2}-\frac{a{T}^2}{4}=aTt-\frac{1}{2}a{t}^2$
$\Rightarrow \frac{T^2}{4}=Tt-\frac{t^2}{2}$
Multiplying by $4$ on both sides
$T^2=4Tt-2t^2$
$\Rightarrow 2t^2-4Tt+T^2=0$
On solving this quadratic equation,
$t=T-\frac{T}{\sqrt{2}}$
$\Rightarrow t=T\left(1-\frac{1}{\sqrt{2}}\right)$
Retardation $\to a$
Initial velocity $\to u$
(I) For total journey
$v=u+at$
$0=u-aT$
$\Rightarrow u=aT$
$d=uT-\frac{1}{2}a{T}^2$
Dividing by 2 on both sides
$\frac{d}{2}=\frac{uT}{2}-\frac{1}{2}\frac{a{T}^2}{2}$ ...(ii)
(II) For half journey
$\frac{d}{2}=ut-\frac{1}{2}a{t}^2$ ...(iii)
On comparing equation (i) $\&$ (iii)
$\frac{uT}{2}-\frac{1}{2}\frac{a{T}^2}{2}=ut-\frac{1}{2}a{t}^2$
Put $u=aT$
$\Rightarrow \frac{a{T}^2}{2}-\frac{a{T}^2}{4}=aTt-\frac{1}{2}a{t}^2$
$\Rightarrow \frac{T^2}{4}=Tt-\frac{t^2}{2}$
Multiplying by $4$ on both sides
$T^2=4Tt-2t^2$
$\Rightarrow 2t^2-4Tt+T^2=0$
On solving this quadratic equation,
$t=T-\frac{T}{\sqrt{2}}$
$\Rightarrow t=T\left(1-\frac{1}{\sqrt{2}}\right)$