Question

A bicyclist is traveling at 2.0m/s then gets faster when traveling down a hill, reaching 19 m/s in only 3.5s. a) What was the acceleration of the rider, assuming it was constant for the ride? b) How long was this hill?

Answer 1

a.
`4.857m/s^2`

b.36.75 m

Step-by-step explanation:

We are given that

Initial speed, u=2m/s

Final speed, v=19 m/s

Time, t=3.5 s

a.

We know that

Acceleration, a=
`(v-u)/(t)`

Using the formula

`a=(19-2)/(3.5)`

`a=(17)/(3.5)`

`a=4.857m/s^2`

Hence, the acceleration of the rider,
`a=4.857m/s^2`

b.

We know that

`v^2-u^2=2as`

Using the formula

`(19)^2-2^2=2* 4.857s`

`361-4=9.714s`

`357=9.714s`

`s=(357)/(9.714)`

`s=36.75 m`

Length of hill=36.75m