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breaghw · Sep 19, 2009, 09:46 AM

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?

The 2 second time change is really throwing me off, how do I work it into the calculation?

Unknown008 · Sep 19, 2009, 10:09 AM

Ok, a graph would really help you here. I'll show you how to do this.

1. Let's say, the cyclist takes 't' seconds to catch up his friend from the time that his friend first met him.

Then, the distance 's' his friend covered until they meet is given by: $s = 3.5 \times t = 3.5t \,m$

The relationship between distance, time and acceleration is given by $s = \frac12 at^2$ (when initial speed is zero)

The acceleration 'a' being 2.4, the displacement 's' being unknown, the time being (t-2)

Now, the distance here are the same, so, $3.5t = \frac12 (2.4)(t-2)^2$

Solve for t to get the time from where they first met. Then, remove 2 from that to have the time that the cyclist takes to reach his friend.

2. Now that you have the time, you can find the distance by replacing 't' in either equations you had above.

3. To find the speed, you can use v = u+at, where v is the final speed, u the initial speed (0 m/s), a the acceleration and t the time the cyclist took.

I hope I helped! :)