Question

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

• A. $20\text{m/s},32\text{m/s}$
• B. $28\text{m/s},36\text{m/s}$
• C. $28\text{m/s},40\text{m/s}$
• D. $24\text{m/s},36\text{m/s}$

Answer 1

The correct option is B $28\text{m/s},36\text{m/s}$

Given, speed of the particle as $v\left(t\right)=6t^2+2t$
So, the speed of the particle at $t=2\text{sec}$, is given as
$v(t=2\text{sec})=6\times 2^2+2\times 2=28\text{m/s}$

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 $v\left(t\right)=6t^2+2t$
and, distance $={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow S={\int }_0^4(6t^2+2t)dt=[2t^3+t^2{]}_0^4$
$\Rightarrow S=(2\times 4^3+4^2)-(0+0)=144\text{m}$
Thus, average speed of the car is $\frac{144}{4}=36\text{m/s}$