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 $t$ is

• A. $2t^4-2t^3+2t\text{m}$
• B. $t^4-t^3+2t\text{m}$
• C. $2t^4-t^3+2t\text{m}$
• D. $2t^4-t^3+2\text{m}$

Answer 1

The correct option is C $2t^4-t^3+2t\text{m}$

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
$\int v\left(t\right)dt$
$\Rightarrow S=\int (8t^3-3t^2+2)dt=2t^4-t^3+2t\text{m}$