Question

If velocity of a particle is $v\left(t\right)=6t^2\text{m/s}$, the average speed of the particle in $6\text{sec}$ is

• A. $60\text{m/s}$
• B. $48\text{m/s}$
• C. $72\text{m/s}$
• D. $90\text{m/s}$

Answer 1

The correct option is C $72\text{m/s}$

Given, $v\left(t\right)=6t^2$, we can say that the particle has not changed its direction once started.
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t\left(6t^2\right)dt$
$\Rightarrow x_f-x_i=2t^3$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=6,x_f=2\times 6^3=432\text{m}$
So, the average speed of the particle is $\frac{432}{6}=72\text{m/s}$