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=(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 (1)/(60 min/h)=0.2 h` is the time taken

Substituting,

`v=(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=(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 (1)/(3600 s/h)=9.7\cdot 10^(-4) h`

So the acceleration is

`a=(28 mph-18 mph)/(9.7\cdot 10^(-4) h)=10,309 mi/h^2`