એક કણ પ્રારંભિક ગતિ u અને પ્રતિપ્રવેગ a સાથે | vedclass

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Question

(b)

Retardation $\rightarrow a$

Initial velocity $\rightarrow u$

$(I)$ For total journey

$v=u+a t$

$0=u-a T$

$\Rightarrow u=a T \ldo

Vedclass · Accepted Answer

$T\left(1-\frac{1}{\sqrt{2}}\right)$

Vedclass · Answer

(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

$\frac{d}{2}=u t-\frac{1}{2} a t^2 \ldots (iii)$

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