A body starts with an initial velocity ‘u’ and moves with a uniform acceleration ‘a’. When the velocity has increased to 5u, the acceleration is reversed in direction but the magnitude remains the same. The body will now return to the starting point with a velocity of                                                          

Given that

Initial velocity = u

Acceleration $=a$

Final velocity = 5u

Displacement = s

We have ${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2a\text{s}$

${\left(\text{5u}\right)}^2-{\left(\text{u}\right)}^2=2a\text{s}$

$24{\text{u}}^{\text{2}}=2a\text{s}$

$\Rightarrow \text{s}=\frac{24{\text{u}}^{\text{2}}}{2a}=\frac{12{\text{u}}^{\text{2}}}{a}\to \left(1\right)$

Now acceleration is reversed

Now initial velocity = u

Acceleration $=-a$

To reach the starting point displacement required $=-\text{s}=\frac{-12{\text{u}}^{\text{2}}}{a}$

Final velocity v =?

We know ${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2a\text{s}$

${\text{v}}^{\text{2}}-{\left(5\text{u}\right)}^2=2\left(-a\right)\times \frac{-12{\text{u}}^{\text{2}}}{a}$

${\text{v}}^{\text{2}}=25{\text{u}}^{\text{2}}+24{\text{u}}^{\text{2}}$

$\therefore \text{v}=\pm 7\text{u}$  We get

$\therefore \text{v}=-7\text{u}$