Question

A particle moves along the x-axis so that at time t≥0 its position is given by x(t)=2t3+3t2−36t+50. What is the total distance traveled by the particle over the time interval 0≤t≤5 ?

Answer 1

505 units

Explanation:

The position function is

`x(t)=2t^3+3t^2+36t+50`

The distance traveled between
`t=0` and
`t=5\ \text{s}` is required so

`x(5)-x(0)=(2* 5^3+3* 5^2+36* 5+50)-(2* 0^3+3* 0^2+36* 0+50)\\\Rightarrow x(5)-x(0)=505\ \text{units}`

The distance traveled by the particle is 505 units.