Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later, the bicyclist hops on his bike and accelerates at 2.11 m/s2 until he catches his friend. a. How much time does it take until he catches his friend?(b) How far has he traveled in this time? (c) What is his speed when he catches up?

Answer 1

a.
`t=3.44s`

b.
`x=12.45m`

c.
`v_f=7.26(m)/(s)`

Step-by-step explanation:

The bicyclist's friend moves with constant speed. So, we have:

`x=vt`

Th bicyclist moves with constant acceleration and starts at rest (
`v_0=0`). So, we have:

`x=v_0t+(at^2)/(2)\\x=(at^2)/(2)`

a. When he catches his friend, both travels the same distance, thus:

`vt=(at^2)/(2)\\t=(2v)/(a)\\t=(2(3.63(m)/(s)))/(2.11(m)/(s^2))\\t=3.44s`

b. We can use any of the distance equations, since both travels the same distance:

`x=vt\\x=3.63(m)/(s)(3.44s)\\x=12.45m`

c. The bicyclist final speed is:

`v_f=v_0+at\\v_f=at\\v_f=2.11(m)/(s^2)(3.44s)\\v_f=7.26(m)/(s)`