Question

As she climbs a hill, a cyclist slows down from 25 mi/hr to 6 mi/hr in 10 seconds. What is her deceleration?

Answer 1

19 mi/hr/s

Explanation:

Declaration = change in speed/time
So deceleration = (25-6)/10
= 1.9 mi/hr/s

Answer 2

`a = -0.84 ms^(-2)`

Explanation:

The working for this solution will use SI units.

First convert the velocities into SI units, then use the appropriate equation of constant acceleration (aka "suvat equation") to calculate the deceleration.

Convert velocities to kmph
`25 mph : 25 × 8/5 = 40 kmph`
`6 mph : 6 × 8/5 = 9.6 kmph`

`8/5` is the conversion factor for mph to kmph .

Convert velocities to SI (`ms^(-1)`)
`(25 mph) 40 kmph : (40 × 10^3)/3600 = 11.11… ms^(-1)`
`(6 mph) 9.6 kmph : (9.6 × 10^3)/3600 = 2.66… ms^(-1)`

Calculate the deceleration
List the quantities that you know:
`s = X`
`u = 11.11… ms^(-1)`
`v = 2.66… ms^(-1)`
`a = ?`
`t = 10 s`

Select the appropriate equation:
`v=u+at`
`⇒ a = (v-u)/t`

Substitute values in and calculate deceleration:
`a = (v-u)/t = (2.66…-11.11…)/10 = -0.84 ms^(-2)`