Chapter 13, Problem 4P

(a)

To determine

To prove: The distance traveled by the object between times $t=a$ and $t=b$ is the area under the graph of speed v between $t=a$ and $t=b$.

Explanation of Solution

Proof:

The formula for distance is,

$\text{Distance}=\text{Speed }\times \text{Time}$

Speed varies over times so speed v is a function time t, the speed at time $t_1$ is $v\left(t_1\right)$

Thus, the area under the curve of a speed function between time a to b determines the distance traveled by object between times $t=a$ and $t=b$.

(b)

To determine

To find: The distance traveled by the car from $t=0$ to $t=5\text{ s}\text{.}$, if the speed of the car is given by the function $v\left(t\right)=6t+0.1t^3\text{ ft/s}\text{.}$

Answer to Problem 4P

The distance travelled by the car is 66.67 ft.

Explanation of Solution

Given:

The speed of the car is,

$v\left(t\right)=6t+0.1t^3\text{ ft/s}\text{.}$

Calculation:

Calculate the area under the curve $v\left(t\right)=6t+0.1t^3\text{ ft/s}\text{.}$, in interval $0\le t\le 5$,

The formula of area is,

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_k\right)}\Delta x$ (1)

$\Delta x$ can be calculated by below formula,

$\Delta x=\frac{b-a}{n}$ (2)

Formula to calculate the value of $x_k$ is,

$x_k=a+k\Delta x$ (3)

Substitute 0 for a and 5 for b in equation (2) to find the value of $\Delta x$,

$\begin{array}{c}\Delta x=\frac{5-0}{n}\\ =\frac{5}{n}\end{array}$

Substitute 0 for a and $\frac{5}{n}$ for $\Delta x$ in equation (3) to find the value of $x_k$,

$\begin{array}{c}{x}_k=0+k\cdot \frac{5}{n}\\ =\frac{5k}{n}\end{array}$

Find the value of $f\left(x_k\right)$ in function $v\left(t\right)$

$\begin{array}{c}f\left(x_k\right)=6x_k+0.1x_k^3\\ =6\left(\frac{5k}{n}\right)+0.1{\left(\frac{5k}{n}\right)}^3\\ =\frac{30k}{n}+\frac{12.5k^3}{n^3}\end{array}$

Substitute $\frac{5}{n}$ for $\Delta x$, $\frac{5k}{n}$ for $x_k$ and $\frac{30k}{n}+\frac{12.5k^3}{n^3}$ for $f\left(x_k\right)$ in equation (1) to find the value of area,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left[\frac{30k}{n}+\frac{2.5k^2}{n^2}\right]}\left(\frac{5}{n}\right)\\ =\underset{n\to \infty }{\lim }\left({\displaystyle \sum _{k=1}^n\frac{150k}{n^2}+\frac{12.5k^2}{n^3}}\right)\\ =\underset{n\to \infty }{\lim }\frac{150}{n^2}{\displaystyle \sum _{k=1}^{n}k}-\underset{n\to \infty }{\lim }\frac{12.5}{n^3}{\displaystyle \sum _{k=1}^n{k}^2}\end{array}$

The value of, ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Substitute $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$, $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ for ${\displaystyle \sum _{k=1}^n{k}^2}$ in the above value of A,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }\frac{150}{n^2}\frac{n\left(n+1\right)}{2}-\underset{n\to \infty }{\lim }\frac{12.5}{n^3}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =75\underset{n\to \infty }{\lim }\frac{n^2+n}{n}-\frac{12.5}{6}\underset{n\to \infty }{\lim }\frac{2n^3+3n^2+n}{n^3}\\ =75\underset{n\to \infty }{\lim }\left(\frac{1}{n}+1\right)-\frac{12.5}{6}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}A=75\cdot 1-\frac{12.5}{6}\left(2\right)\\ =66.67\text{ ft}\text{.}\end{array}$

The distance traveled by the car is equal to the area A.

Thus, the distance travelled by the car is 66.67 ft.