# Finding the displacement and the distance traveled by a particle from the velocity function

The velocity function (in meters per second) for a particle moving along a line is given by v(t)=t^3−3t^2. Find the displacement and the distance traveled by the particle during the time interval [-1,5]

To find the displacement we simply integrate the velocity function
$\text{Displacement}=\int_{-1}^{5}t^3-3t^2dt=\frac{t^4}{4}-t^3 |_{-1}^{5}$
$=(\frac{5^4}{4}-5^3)-(\frac{1}{4}+1)=\frac{5^4-4\cdot 5^3-5}{4}=\frac{120}{4}=30.$

To find the distance traveled we should integrate absolute value of the velocity function:
$\text{Distance traveled}=\int_{-1}^{5}|t^3-3t^2|dt.$
We have
$t^3-3t^2=t^2(t-3).$
Hence $|t^3-3t^2|=t^2 (t-3)=t^3-3t^2 \text{for} t>3$, and $|t^3-3t^2|=-t^2 (t-3)=-t^3+3t^2 \text{for} t<3.$ Thus
$\text{Distance traveled}=\int_{-1}^{5}|t^3-3t^2|dt=\int_{-1}^{3}-t^3+3t^2dt+\int_3^5t^3-3t^2 dt$
$=(-\frac{t^4}{4}+t^3)\big|_{-1}^{3}+(\frac{t^4}{4}-t^3)\big|_3^5$
$=(-\frac{3^4}{4}+3^3)-(-\frac{1}{4}-1)+(\frac{5^4}{4}-5^3)-(\frac{3^4}{4}-3^3)$
$=2(-\frac{3^4}{4}+3^3)+(\frac{1}{4}+1)+(\frac{5^4}{4}-5^3)$
$=2(-\frac{81}{4}+27)+(\frac{1}{4}+1)+(\frac{625}{4}-125)=46.$

• Savionf
+1

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