Question

I traveled 3.8 miles in 12 minutes. Calculate my average velocity in miles per hour. 2) A sprinter increased his/her speed from 18 mph to 28 mph over 3.5 seconds. Calculate his/her acceleration.

Answer 1

1) Average velocity: 19 mi/h

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

In this problem,

d = 3.8 mi is the displacement

$t=12 min \cdot \frac{1}{60 min/h}=0.2 h$ is the time taken

Substituting,

$v=\frac{3.8 mi}{0.2 h}=19 mi/h$

Note that this is actually the average speed, not the average velocity, since there is no information about the direction of the motion.

2) $10,309 mi/h^2$

The acceleration is given by

$a=\frac{v-u}{t}$

where we have

v = 28 mph is the final velocity

u = 18 mph is the initial velocity

t = 3.5 s is the time taken

Conerting the time from seconds to hours,

$t=3.5 s \cdot \frac{1}{3600 s/h}=9.7\cdot 10^{-4} h$

So the acceleration is

$a=\frac{28 mph-18 mph}{9.7\cdot 10^{-4} h}=10,309 mi/h^2$