Question

If the speed of the particle is $v=8t^3-3t^2+2\text{m/s}$, the distance of the particle as a function of time $t$ and the average speed in $2\text{sec}$ respectively are

• A. $2t^4-t^3+2t\text{m},14\text{m/s}$
• B. $2t^4-t^3+t\text{m},14\text{m/s}$
• C. $2t^4-t^3+2t\text{m},28\text{m/s}$
• D. $t^4-t^3+2t\text{m},14\text{m/s}$

Answer 1

The correct option is A $2t^4-t^3+2t\text{m},14\text{m/s}$

Given, the speed of the particle is $v\left(t\right)=8t^3-3t^2+2$
So, the distance as a function of time is given as
$\int v\left(t\right)dt$
$\Rightarrow S=\int (8t^3-3t^2+2)dt=2t^4-t^3+2t\text{m}$
Thus, the distance travelled in $2\text{sec}$ is $S=2\times 2^4-2^3+2\times 2=28\text{m}$
Hence, average speed of the particle is
$\frac{\text{Total distance travelled (S)}}{\text{Total time taken (t)}}=\frac{28}{2}=14\text{m/s}$