Chapter 5, Problem 63RE

To determine

To find: the displacement and the distance traveled by the particle during the time interval [0, 5].

Answer to Problem 63RE

The displacement is equal to $\frac{175}{6}$ .

The distance is equal to $\frac{59}{2}$ .

Explanation of Solution

Given information: A particle moves along a line with velocity function $v(t)=t^2-t$ ,

where v is measured in meters per second.

Formula used:

  ${\displaystyle \int x^a}dx=\frac{1}{a+1}{x}^{a+1}$ .

Calculation:

The displacement is the integral of the velocity.

  $\begin{array}{c}s(t)={\displaystyle \int v(t)dt={\displaystyle \underset{0}{\overset{5}{\int }}(t^2}}-t)dt\\ =\left[\frac{1}{3}{t}^3-\frac{1}{2}{t}^2\right]\begin{array}{c}5\\ 0\end{array}\\ =\frac{1}{3}{5}^3-\frac{1}{2}{5}^2-(\frac{1}{3}{0}^3-\frac{1}{2}{0}^2)\\ =\frac{125}{3}-\frac{25}{2}\\ =\frac{175}{6}\end{array}$

The distance is equal to the integral of the absolute value of the velocity.

  $\text{Distance=}{\displaystyle \underset{0}{\overset{5}{\int }}\left|v(t)\right|}dt$ .

  

From 0< x <1 v(t) is negative so the integral from 0 to 1 has to be negative.

  $\begin{array}{c}\text{Distance=}{\displaystyle \underset{0}{\overset{5}{\int }}\left|v(t)\right|}dt=-{\displaystyle \underset{0}{\overset{1}{\int }}v(t)dt+{\displaystyle \underset{1}{\overset{5}{\int }}v(t)dt}}\\ =-{\displaystyle \underset{0}{\overset{1}{\int }}(t^2-t)dt+{\displaystyle \underset{1}{\overset{5}{\int }}(t^2-t)dt}}\\ =-\left[\frac{1}{3}{t}^3-\frac{1}{2}{t}^2\right]\begin{array}{c}1\\ 0\end{array}+\left[\frac{1}{3}{t}^3-\frac{1}{2}{t}^2\right]\begin{array}{c}5\\ 1\end{array}\\ =-\left[(\frac{1}{3}{1}^3-\frac{1}{2}{1}^2)-(\frac{1}{3}{0}^3-\frac{1}{2}{0}^2)\right]+\left[(\frac{1}{3}{5}^3-\frac{1}{2}{5}^2)-(\frac{1}{3}{1}^3-\frac{1}{2}{1}^2)\right]\\ =\frac{59}{2}\end{array}$