Question

If the speed of the particle is $v=4t^3+6\text{m/s}$, the distance of the particle as a function of time $t$ and the average speed in $3\text{sec}$ respectively are

• A. $t^4-6t\text{m},33\text{m/s}$
• B. $2t^4-6t\text{m},21\text{m/s}$
• C. $t^4+6t\text{m},33\text{m/s}$
• D. $t^3-6t\text{m},21\text{m/s}$

Answer 1

The correct option is C $t^4+6t\text{m},33\text{m/s}$

Given, the speed of the particle is $v\left(t\right)=4t^3+6$
So, the distance as a function of time is given as
$\int v\left(t\right)dt$
$\Rightarrow S=\int (4t^3+6)dt=t^4+6t\text{m}$
Thus, the distance travelled in $3\text{sec}$ is $S=3^4+6\times 3=99\text{m}$
Hence, average speed of the particle is
$\frac{\text{Total distance travelled (S)}}{\text{Total time taken (t)}}=\frac{99}{3}=33\text{m/s}$