[springraindrops]

OK, I've got a cyclist who rides by at a constant speed of 3.5 m/s and 2 seconds later a second cyclist hops on his bike and accelerates at 2.4 m/s until he catches his friend. How much time does it take for him to catch his friend? How har has he traveled in this time? What is his speed when he catches up?

Help! I am totally lost as to where to start! I need the answer by tomorrow!

[eeseely]

PROBLEM

OK, I've got a cyclist who rides by at a constant speed of 3.5 m/s and 2 seconds later a second cyclist hops on his bike and accelerates at 2.4 m/s until he catches his friend. How much time does it take for him to catch his friend? How har has he traveled in this time? What is his speed when he catches up?

Given:

A: v = 3.5m/s
B: a = 2.4m/s/s
B starts 2 seconds after A started

Find:

t (time) for B to get even with A
B's distance traveled = s
B's velocity at that time = v

To solve this problem, set up a relationship where the distance travelled
by both A and B is equal

Hence:

3.5m/sec * 2s + 3.5t = 2.4 m/s/s*(t ^2)

7.0m + (3.5t) = 2.4m/s/s*(t sqd) ( NOTE: at t = 0, A has travelled 7.0m)

2.4t^2 sqd - 3.5t –7 = 0

Use the quadratic equation to solve for t

A = +2.4
B = -3.5
C = -7.0

Once you fint t you can solve for s
and for B's v