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:

The particle starts from rest.

$\therefore$u = 0

Uniform acceleration = α
For t = t₁
We get, 1$\text{v}=\mathrm{\alpha }{\text{t}}_{\text{1}}$

Travelled distance =x

We get, $x=\frac{1}{2}\alpha {\text{t}}_{\text{1}}^{\text{2}}$    ______(1)

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

u = αt₁ , acceleration=α

v=0 , time taken =t₂

We get v=αt₁=βt₂ _________(2)

The distance travelled by it is (s−x)  where s is the total distance travelled

$\text{s−x=}\frac{1}{2}{{\mathrm{\beta t}}_2}^2$    ______________(3)

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 $\alpha =\frac{\text{v}}{{\text{t}}_{\text{1}}},\text{\hspace{0.17em}β}=\frac{\text{v}}{{\text{t}}_{\text{2}}}$           

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

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

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

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