Answers:
a) [tex]0.263 mi/min[/tex]
b) [tex]15.789 mi/h[/tex]
Explanation:
a) We are told Steve Cram ran a distance of 1 mile ([tex]d=1 mi[/tex]) in a time of [tex]t=3.81 min[/tex]. As velocity [tex]V[/tex] is defined as:
[tex]V=\frac{d}{t}[/tex]
Steve's velocity in [tex]mi/min[/tex] was:
[tex]V=\frac{1 mi}{3.81 min}=0.263 \frac{mi}{min}[/tex]
b) Now, if we want to express thi same velocity in miles per hour ([tex]mi/h[/tex]), we have to make the conversion, taking into account [tex]1h=60 min[/tex]:
[tex]V=0.263 \frac{mi}{min} \frac{60 min}{1 h}[/tex]
[tex]V=15.789 \frac{mi}{h}[/tex]
The average speed should be
a. 0.263 mi/min
b. 15.789 mi.hr
a. The speed in mi/min should be
[tex]= 1 mi \div 3.81[/tex]
= 0.263 mi/min
b. The speed in mi/hr should be
[tex]= 0.263 \times 60 min \div 1 hour\\\\[/tex]
= 15.789 mi.hr
Here we considered one hour = 60 minutes
learn more about the speed here: https://brainly.com/question/17894094
