Application of the equations of motion with constant acceleration

[sillybean]

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

Homework Statement

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?

Homework Equations

V=Vo + at

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

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

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 + volt + 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 + volt + 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?

 

[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 traveled at a const speed for t secs, whereas the repairing guy has traveled the same dist in (t-2) s. The rest you know.

 

[physixguru]

Homework Statement

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?

Homework Equations

V=Vo + at

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

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

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 + volt + 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 + volt + 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?

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

 

[sillybean]

thanks again you guys are life savers