Question

A body starts from a point O with velocity v and acceleration a. The direction of acceleration is reversed when the velocity becomes 5v. The velocity of the body at point O will be

• A. -v
• B. -5v
• C. -7v
• D. -9v

Answer 1

The correct option is C

-7v

Step 1: Given data:

1. A body starts from a point O with velocity v and acceleration a.
2. The direction of acceleration is reversed when the velocity becomes 5v

Now,

Velocity at O = $vm/s$

Acceleration at O =$am/s^2$

Velocity when the direction is reversed= $5vm/s$

Acceleration when the direction is reversed will be= $-am/s^2$

Step 2: From equations of motion we have:

$v^2-u^2=2as$

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

Step 3: Calculating the distance travelled (with positive acceleration) s:

$$\begin{gathered} {\left(5v\right)}^2-v^2=2as \\ s=12\frac{v^2}{a} \end{gathered}$$

Let the distance travelled before the body stops due to negative acceleration be s'.

Step 4: Calculating s':

$$\begin{gathered} v=0 \\ u=5v \\ a=-a \end{gathered}$$

So,

$$\begin{gathered} 0-{\left(5v\right)}^2=2(-a)s\text{'} \\ -{\left(5v\right)}^2=-2as\text{'} \\ s\text{'}=25\frac{v^2}{2a} \end{gathered}$$

So, total distance travelled = $s+s\text{'}$ = $49\frac{v^2}{2a}$

Step 5: Calculating Final Velocity:

Let the final velocity at O be v'

When the body is stopped due to negative acceleration and then starts moving in the opposite direction, its initial velocity will be zero and the distance it will cover to return to O will be $\frac{49v^2}{2a}$

Hence, the final velocity is -7v.

Therefore, the correct option is (C).