# One dimension kinetics

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5m/s. Two seconds later the bicylicts hops on his bike and accelerates at 2.4m/s^2 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?

 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #2 Feb 7, 2007, 12:10 AM
This should be fairly easy using suvat equations.

In a) you need an equation for s for both cyclists, equal them to each other, and solve for t.

For the other parts you are simply doing the same, making different equations for each cyclist, making them equal and solving for the number you are looking for.

Let me know if you need me to go deeper than that.
 Sduduzo Posts: 2, Reputation: 10 Junior Member #3 Feb 9, 2012, 05:07 AM
If sound travels at a constant speed of 343m/s in air, how much time (in seconds) does it take for the sound of thunder to travel 1609 m?
 Sduduzo Posts: 2, Reputation: 10 Junior Member #4 Feb 9, 2012, 05:18 AM
(a)
accelaration=change in velocity/change in time
2.4 m.s-2=3.5/change in time
change in time=1.458s