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  4. 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

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Q. 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

Motion in a Straight Line

Solution:

image
Retardation $\rightarrow a$
Initial velocity $\rightarrow u$
(I) For total journey
$v =u +a t$
$0 =u-a T$
$\Rightarrow u =a T$
$d=u T-\frac{1}{2} a T^{2}$
Dividing by 2 on both sides
$\frac{d}{2}=\frac{u T}{2}-\frac{1}{2} \frac{a T^{2}}{2}$ ...(ii)
(II) For half journey
$\frac{d}{2}=u t-\frac{1}{2} a t^{2}$ ...(iii)
On comparing equation (i) $\&$ (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,
$t=T-\frac{T}{\sqrt{2}}$
$\Rightarrow t=T\left(1-\frac{1}{\sqrt{2}}\right)$image
Retardation $\rightarrow a$
Initial velocity $\rightarrow u$
(I) For total journey
$v =u +a t$
$0 =u-a T$
$\Rightarrow u =a T$
$d=u T-\frac{1}{2} a T^{2}$
Dividing by 2 on both sides
$\frac{d}{2}=\frac{u T}{2}-\frac{1}{2} \frac{a T^{2}}{2}$ ...(ii)
(II) For half journey
$\frac{d}{2}=u t-\frac{1}{2} a t^{2}$ ...(iii)
On comparing equation (i) $\&$ (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,
$t=T-\frac{T}{\sqrt{2}}$
$\Rightarrow t=T\left(1-\frac{1}{\sqrt{2}}\right)$