# Application of the equations of motion with constant acceleration

1. Jan 23, 2008

### sillybean

[SOLVED] Application of the equations of motion with constant acceleration

1. The problem statement, all variables and given/known data
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 2.2 m/s^2 until he catches his friend.

a)How much time does it take until he catches his friend (after his friend passes him)?
b)How far has he traveled in this time?
c)What is his speed when he catches up?

2. Relevant equations

V=Vo + at

X= Xo + Vot + 1/2at^2

X= Xo + 1/2(Vo + V)t

3. The attempt at a solution
Okay so I tried figuring out the distance that the friend who passed the one doing repairs traveled in 2 seconds. Since his velocity was constant, I reasoned that he traveled 8 meters in 2 seconds. Then I plugged in the acceleration given in the problem along with the distance i figured into the equation: X= Xo + Vot + 1/2at^2. (I made Xo and Vo = zero since the guy doing the repairs starts from rest)

It looked like this: 8.0m = 0+0+ 1/2(2.2m/s^2)t^2
8.0m = 1.1m/s^2(t^2)
7.27s^2=t^2 (took square root of both sides)
t=2.7s

but that was wrong.

I then figured the distance I reasoned was wrong and actually used an equation this time.
I used X= Xo + 1/2(Vo + V)t
X = 0 + 1/2 (4.0 m/s)(2.0s)
X = 4.0 m

I redid my work using this new distance (which I already thought was wrong >:C)
X= Xo + Vot + 1/2at^2
4.0m = 0 + 0 + 1/2(2.2)t^2
4.0 = 1.1t^2
3.6 = t^2 (square root)
t=1.9s

Guess what. It was wrong.

What am I doing wrong?

2. Jan 24, 2008

### Shooting Star

Suppose it has taken a total of t secs from the time when the friend passes him by to the time when he catches up. Then the friend has travelled at a const speed for t secs, whereas the repairing guy has travelled the same dist in (t-2) s. The rest you know.

3. Jan 24, 2008

### physixguru

use concepts of relative velocity...this shall make the problem much easier...

4. Jan 28, 2008

### sillybean

thanks again you guys are life savers