Question

A body starts moving with a velocity $v_0=10m/s$. It experiences retardation equal to $0.2v^2$. Its velocity after $2s$ is given by

• A. $+2m/s$
• B. $+4m/s$
• C. $-2m/s$
• D. $+6m/s$

Answer 1

The correct option is A

$+2m/s$

Step 1: Given data

1. Acceleration (negative Retardation ) of the body is, $a=-0.2v^2$.
2. The initial velocity of the body is $v_0=10m/s$.
3. Given time to travel = t seconds.

Step 2: Velocity

1. Velocity is the time rate of change in displacement,i.e, $v=\frac{s}{t}$, where s is the displacement and t is the time.
2. Again the time rate of change in velocity is called acceleration, i.e, acceleration, $a=\frac{dv}{dt}$, where v is the velocity and t is the time.

Step 3: Finding the velocity

According to the question, acceleration, $a=-0.2v^2$.

$$\begin{gathered} \frac{dv}{dt}=-0.2v^2 \\ or-\frac{dv}{o.2v^2}=dt..............\left(1\right) \end{gathered}$$

Integrating equation (1) we get,

$$\begin{gathered} -5{\int }_{10}^V\frac{dv}{v^2}={\int }_0^{2}dt \\ or-5{\left[-\frac{1}{v}\right]}_{10}^V=\left[t\right]{}_0^2 \\ or5\left(\frac{1}{v}-\frac{1}{10}\right)=2 \\ or\frac{5}{v}-\frac{5}{10}=2 \\ or\frac{5}{v}=2+\frac{1}{2}=\frac{5}{2} \\ or5v=10 \\ orv=2 \end{gathered}$$

So, the velocity of the body is $2m/s$. Therefore, option (A) is correct.