Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 4.0 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.3 m/s2 until he catches his friend.

Part A

How much time does it take until he catches his friend (after his friend passes him)?

Express your answer using two significant figures.

t =

  s  

Part B

How far has he traveled in this time?

Express your answer using two significant figures.

x =

  m  

Part C

What is his speed when he catches up?

Express your answer using two significant figures.

Answer 1

A)

let time taken is t

distance moved by friend = speed * time

= 4.0*t

distance moved by bicyclist = 0.5*a*time^2

= 0.5*2.3*(t-2)^2

= 1.15*(t-2)^2

equate both:

1.15*(t-2)^2 = 4.0*t

1.15*(t^2 - 4t + 4) = 4.0*t

1.15*t^2 - 4.6*t + 4.6 = 4.0*t

1.15*t^2 - 8.6*t + 4.6 = 0

This is quadratic equation (at^2+bt+c=0)

a = 1.15

b = -8.6

c = 4.6

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 52.8

roots are :

t = 6.9 and t = 0.58

t can't be 0.58 as this will make t-2 negative

t = 6.9

Answer: 6.9 s

B)

DISTANCE TRAVELLED BU BOTH IS SAME

use:

d = 4.0*t

= 4.0* 6.9

= 28 m

Answer: 28 m

C)

v = a*(t-2)

= 2.3*(6.9-2)

= 11 m/s

Answer: 11 m/s