Chapter 1, Problem 67SGA

To determine

To find: the distance traveled at the end of each second for 5 seconds.

Answer to Problem 67SGA

  $\begin{array}{l}{d}_1=10\text{ m}\\ d_2=40\text{ m}\\ d_3=90\text{ m}\\ d_4=160\text{ m}\\ d_5=250\text{ m}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}d=\frac{1}{2}a{t}^2\\ a=20{\text{ m/s}}^2\\ t_1=1\sec \\ t_2=2\sec \\ t_3=3\sec \\ t_4=4\sec \\ t_5=5\sec \end{array}$

Calculation:

Substitute the value of a and $t_1$ to calculate the distance as follows:

  $\begin{array}{c}d=\frac{1}{2}a{\left(t_1\right)}^2\\ d=\frac{1}{2}\left(20\right){\left(1\right)}^2\\ =10\text{ m}\end{array}$

Thus, distance travelled at 1 second is 10 m.

Substitute the value of a and $t_2$ to calculate the distance as follows:

  $\begin{array}{c}d=\frac{1}{2}a{\left(t_2\right)}^2\\ d=\frac{1}{2}\left(20\right){\left(2\right)}^2\\ =40\text{ m}\end{array}$

Thus, distance travelled at 2 second is 40 m.

Substitute the value of a and $t_3$ to calculate the distance as follows:

  $\begin{array}{c}d=\frac{1}{2}a{\left(t_3\right)}^2\\ d=\frac{1}{2}\left(20\right){\left(3\right)}^2\\ =90\text{ m}\end{array}$

Thus, distance travelled at 3 second is 90 m.

Substitute the value of a and $t_4$ to calculate the distance as follows:

  $\begin{array}{c}d=\frac{1}{2}a{\left(t_4\right)}^2\\ d=\frac{1}{2}\left(20\right){\left(4\right)}^2\\ =160\text{ m}\end{array}$

Thus, distance travelled at 4 second is 160 m.

Substitute the value of a and $t_5$ to calculate the distance as follows:

  $\begin{array}{c}d=\frac{1}{2}a{\left(t_5\right)}^2\\ d=\frac{1}{2}\left(20\right){\left(5\right)}^2\\ =250\text{ m}\end{array}$

Thus, distance travelled at 5 second is 250 m.

To determine

To find: is this relation a function and give explanation.

Answer to Problem 67SGA

Yes

Explanation of Solution

Given information:

d (meter)	t (sec)
10	1
40	2
90	3
160	4
250	5

Calculation:

Yes, it’s a function because each element of domain is paired with exactly one element of the range.