hw 6 - average rate of change section 2.1: problem 6 (1 point)\nthe table below shows the position of a…

hw 6 - average rate of change section 2.1: problem 6 (1 point)\nthe table below shows the position of a cyclist. find the average velocity for each time period.\n| t (in seconds) | 0 | 1 | 2 | 3 | 4 | 5 |\n| s (in meters) | 0 | 1.5 | 5.4 | 10.3 | 17.7 | 25.6 |\n1. 1,3\naverage velocity: m/s\n2. 2,3\naverage velocity: m/s\n3. 3,5\naverage velocity: m/s\n4. 3,4\naverage velocity: m/s

hw 6 - average rate of change section 2.1: problem 6 (1 point)\nthe table below shows the position of a cyclist. find the average velocity for each time period.\n| t (in seconds) | 0 | 1 | 2 | 3 | 4 | 5 |\n| s (in meters) | 0 | 1.5 | 5.4 | 10.3 | 17.7 | 25.6 |\n1. 1,3\naverage velocity: m/s\n2. 2,3\naverage velocity: m/s\n3. 3,5\naverage velocity: m/s\n4. 3,4\naverage velocity: m/s

Answer

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