A particle starting from rest, reaches a maximum velocity ‘v’ with a uniform acceleration and comes to rest with a uniform deceleration by covering a total distance ‘s’ moving along a straight line. The total time of motion is

Given that

A particle

u = 0

Uniform acceleration $=\alpha$

For $\text{t}={\text{t}}_{\text{1}}$

We get $\text{v}=\alpha {\text{t}}_{\text{1}}$

Travelled distance $=x$

We get $x=\frac{1}{2}\alpha {\text{t}}_{\text{1}}^{\text{2}}\to \left(1\right)$

Later it started retarding for ${\text{t}}_{\text{2}}$  and come to rest

$\text{u}=\alpha {\text{t}}_{\text{1}},acceleration=\alpha$

$\text{v}=0,\text{time}\text{taken}={\text{t}}_{\text{2}}$

We get $\text{v}=\alpha {\text{t}}_{\text{1}}=\beta {\text{t}}_{\text{2}}\to \left(2\right)$

The distance travelled by it is $\left(\text{s}-x\right)$  where s is the total distance travelled

$\text{s}-x=\frac{1}{2}\beta {\text{t}}_{\text{2}}^{\text{2}}\to \left(3\right)$

We get from (1) & (3)                         

$\text{s}=\frac{1}{2}\alpha {\text{t}}_{\text{1}}^{\text{2}}+\frac{1}{2}\beta {\text{t}}_{\text{2}}^{\text{2}}$                                 From (2) we get $\alpha =\frac{\text{v}}{{\text{t}}_{\text{1}}},\beta =\frac{\text{v}}{{\text{t}}_{\text{2}}}$           

$\text{s}=\frac{1}{2}\frac{\text{v}}{{\text{t}}_{\text{1}}}\cdot {\text{t}}_{\text{1}}^{\text{2}}+\frac{1}{2}\frac{\text{v}}{{\text{t}}_{\text{2}}}\cdot {\text{t}}_{\text{2}}^{\text{2}}$

$\text{s}=\frac{1}{2}{\text{vt}}_{\text{1}}+\frac{1}{2}{\text{vt}}_{\text{2}}$

$\Rightarrow \text{s}=\frac{\text{v}}{\text{2}}\left({\text{t}}_{\text{1}}+{\text{t}}_{\text{2}}\right)\text{let}\text{t}={\text{t}}_{\text{1}}+{\text{t}}_{\text{2}}$

$\Rightarrow 2\text{s}=\text{vt}$

$\therefore \text{t}=\frac{\text{2s}}{\text{v}}$