Question

hw 6 - average rate of change section 2.1: problem 6 (1 point)
the table below shows the position of a cyclist. find the average velocity for each time period.

t (in seconds)	0	1	2	3	4	5
s (in meters)	0	1.5	5.4	10.3	17.7	25.6

1. 1,3

average velocity: m/s

1. 2,3

average velocity: m/s

1. 3,5

average velocity: m/s

1. 3,4

average velocity: m/s

Explanation:

Step1: Recall average - velocity formula

The formula for average velocity $v_{avg}$ over the interval $[t_1,t_2]$ is $v_{avg}=\frac{s(t_2)-s(t_1)}{t_2 - t_1}$, where $s(t)$ is the position - function.

Step2: Calculate for $[1,3]$

We have $t_1 = 1$, $t_2=3$, $s(1)=1.5$, and $s(3)=10.3$. Then $v_{avg}=\frac{s(3)-s(1)}{3 - 1}=\frac{10.3 - 1.5}{2}=\frac{8.8}{2}=4.4$ m/s.

Step3: Calculate for $[2,3]$

We have $t_1 = 2$, $t_2=3$, $s(2)=5.4$, and $s(3)=10.3$. Then $v_{avg}=\frac{s(3)-s(2)}{3 - 2}=\frac{10.3 - 5.4}{1}=4.9$ m/s.

Step4: Calculate for $[3,5]$

We have $t_1 = 3$, $t_2=5$, $s(3)=10.3$, and $s(5)=25.6$. Then $v_{avg}=\frac{s(5)-s(3)}{5 - 3}=\frac{25.6 - 10.3}{2}=\frac{15.3}{2}=7.65$ m/s.

Step5: Calculate for $[3,4]$

We have $t_1 = 3$, $t_2=4$, $s(3)=10.3$, and $s(4)=17.7$. Then $v_{avg}=\frac{s(4)-s(3)}{4 - 3}=\frac{17.7 - 10.3}{1}=7.4$ m/s.

Answer:

1. $4.4$ m/s
2. $4.9$ m/s
3. $7.65$ m/s
4. $7.4$ m/s