Question

the velocity function, in feet per second, is given for a particle moving along a straight line, where t is the time in seconds. Findthe displacement andthe total distance that the particle travels over the given interval.

Answer 1

When given a displacement function, its derivative is the velocity function. Vice versa, when given a velocity function, its integral is the displacement function.

So, we need to integrate the given function to find the displacement function.

$\begin{gathered} \int(t^3-8t^2+15t)dt \\ \\ =\int t^3dt-\int8t^2dt+\int15tdt \\ \\ =\int t^3dt-8\int t^2dt+15\int tdt \\ \\ =\frac{t^4}{4}-8(\frac{t^3}{3})+15(\frac{t^2}{2})+C \\ \\ =\frac{t^4}{4}-\frac{8t^3}{3}+\frac{15t^2}{2}+C \end{gathered}$

To solve for C, we need to have a known displacement. We know that when t = 0, the object has not moved yet. Therefore, it is safe to assume that d must be 0.

$\begin{gathered} \frac{t^4}{4}-\frac{8t^3}{3}+\frac{15t^2}{2}+C=0 \\ \\ \frac{0^4}{4}-\frac{8(0^3)}{3}+\frac{15(0^2)}{2}+C=0 \\ \\ C=0 \end{gathered}$

So now we know that the displacemnet function is:

$d(t)=\frac{t^{4}}{4}-\frac{8t^{3}}{3}+\frac{15t^{2}}{2}$

We can now solve for the total displacement when t = 5.

$\begin{gathered} d(t)=\frac{t^{4}}{4}-\frac{8t^{3}}{3}+\frac{15t^{2}}{2} \\ \\ d(5)=\frac{5^4}{4}-\frac{8(5^3)}{3}+\frac{15(5^2)}{2} \\ \\ d(5)=\frac{625}{4}+\frac{1,000}{3}+\frac{375}{2} \\ \\ d(5)=\frac{8,125}{12}\approx677.083 \end{gathered}$

Therefore, the total displacement is 677.083 feet.

Because the particle is moving along a straight line, then the distance is equal to the displacement.