Question

Micheal and his younger sister Elizabeth are having a race in the backyard. Michael gives his sister a head start and they run for 20 secs. The distance they are along in the race, in feet, is given by; Micheals distance given by the function m(t) and Elizabeth’s distance by the function e(t). Explain your answers1) how far does Michael run during the 20 second race? How far does Elizabeth run?2) how fast do both Michael and Elizabeth travel? (How many ft per second) express answers in decimal form with units

Answer 1

Given,

The graph of the function m(t) and e(t) is shown in the question.

a)From graph it is clearly seen that,

The distance covered by michael in 20 seconds is 50 feet.

The distance covered by elizabeth in 20 seconds is (60-25)= 35 feet.

b)The speed of Michael is,

`\begin{gathered} \text{Speed}=\frac{final\text{ position-initial position}}{final\text{ time-initial time}} \\ =(50-0)/(20-0) \\ =2.5\text{ fe}et\text{ per second} \end{gathered}`

The speed of elizabeth is,

`\begin{gathered} \text{Speed}=\frac{final\text{ position-initial position}}{final\text{ time-initial time}} \\ =(60-25)/(20-0) \\ =1.75\text{ fe}et\text{ per second} \end{gathered}`

Hence, the speed of michael is 2.5 ft/second and elizabeth is 1.75 ft/second.