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

Answer:

505 units

Step-by-step 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

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

The distance traveled by the particle is 505 units.