A bicyclist is finishing his repair on a flat tire when a friend rides by with a constant speed of 3.5m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.4m/s/s 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?
2006-10-12 15:48:32 · 1 answers · asked by hbpq83 1 in Science & Mathematics ➔ Physics
It looks like no one else is going to help you with this so I will take a stab at it.
velocity=acceleration X time
distance=1/2 acceleration X time^2
distance=velocity X time
So: 3.5t+7=1/2 X2.4t^2 (since the two distances have to be equal when they meet, right?)
7t+14=2.4t^2
0=2.4t^2-7t-14
t=(7+-sqrt(49+4X2.4X14))/4.8
t~=4.28...seconds
distance~=3.5 X 4.28 meters~=14.98 Meters
speed~=2.4 X 4.28 meters/sec~=10.27meters/sec
I believe these answers are approximately correct.
2006-10-12 16:37:13 · answer #1 · answered by Sciencenut 7 · 0⤊ 0⤋