Question

If the speed of a particle is given as $v=3t^2+5\text{m/s}$, the speed of the particle at $4\text{sec}$ and the average speed in $4\text{sec}$ respectively is

• A. $53\text{m/s},84\text{m/s}$
• B. $53\text{m/s},21\text{m/s}$
• C. $50\text{m/s},21\text{m/s}$
• D. $45\text{m/s},20\text{m/s}$

Answer 1

The correct option is B $53\text{m/s},21\text{m/s}$

Given, speed of the particle as $v\left(t\right)=3t^2+5$
So, the speed of the particle at $t=4\text{sec}$, is given as
$v(t=4\text{sec})=3\times 4^2+5=53\text{m/s}$

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 $v\left(t\right)=3t^2+5$
and, distance $={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow S={\int }_0^4(3t^2+5)dt=[t^3+5t{]}_0^4$
$\Rightarrow S=(4^3+5\times 4)-(0+0)=84\text{m}$
Thus, average speed of the car is $\frac{84}{4}=21\text{m/s}$