Question

Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many meters has it traveled during the first t seconds?

Answer 1

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

$v(t)=t^2e^{-2t}\ m/s$

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

$x=\int\limits {v.dt}$

$x=\int\limits {t^2e^{-2t}dt}$

Using by parts integration, we get the value of x as :

$x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters$

Hence, this is the required solution.