Question

If the speed of a particle is given as $v=6t^2+2t+5\text{m/s}$, the speed of the particle at $4\text{sec}$ and the average speed in $4\text{sec}$ respectively is

• A. $100\text{m/s},40\text{m/s}$
• B. $120\text{m/s},50\text{m/s}$
• C. $109\text{m/s},41\text{m/s}$
• D. $109\text{m/s},164\text{m/s}$

Answer 1

The correct option is C $109\text{m/s},41\text{m/s}$

Given, speed of the particle as $v\left(t\right)=6t^2+2t+5$
So, the speed of the particle at $t=4\text{sec}$, is given as
$v(t=4\text{sec})=6\times 4^2+2\times 4+5=109\text{m/s}$

Now, average speed of the car is given by
$\frac{\text{Total distance covered (S)}}{\text{Total time taken (t)}}$
Given, the speed of the car is $v\left(t\right)=6t^2+2t+5$
and, distance $={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow S={\int }_0^4(6t^2+2t+5)dt=[2t^3+t^2+5t{]}_0^4$
$\Rightarrow S=(2\times 4^3+4^2+5\times 4)-(0+0)=164\text{m}$
Thus, average speed of the car is $\frac{164}{4}=41\text{m/s}$