Question

A particle is moving along a straight line such that square of its velocity varies with time as shown in figure. What is the acceleration of the particle at t = 4s?

• A. 4 $m/s^2$
• B. 1/4 $m/s^2$
• C. 1/2 $m/s^2$
• D. 0

Answer 1

The correct option is B 1/4 $m/s^2$

The equation of a line passing through origin is like $y=mx$ where $m$ is slope.

Similarly in given graph $v^2$ is on $y$ axis and $t$ on $x$ axis so equation becomes $v^2=mt$

Slope $m=(4-0)/(4-0)=1$

So equation becomes $v^2=t$ or $v=\sqrt[2]{2t}$...........(1)

now to find the acceleration just differentiate the equation with respect to $t$ we get $2v\frac{dv}{dt}=1$ but $dv/dt$ is acceleration $a$ so the equation-1 becomes $2va=1$ or $a=\frac{1}{2v}=\frac{1}{2\sqrt[2]{2t}}$

Now put $t=4s$ we get $a=\frac{1}{4}m/s^2$

Option B is correct.