Kinematics bus problem

1. Nov 26, 2009

lovemake1

1. The problem statement, all variables and given/known data
Question reads: A bus travelling at 15.0m/s [E], started to slow down with an acceleration of 0.5m/s^2[W] as soon as it was abreast of a sports car. At the same instant, the sports car had a velocity of 6m/s [E] and was accelerating at 2.5m/s^2 [E]. At what time does the sports car catch up to the bus?

2. Relevant equations
d = Vit + 1/2at^2

3. The attempt at a solution

d = 15t + 1/2(-0.5)t^2 <-- bus
d = 6t + 1/2(2.5)t^2 <-- car

these are the two equations that i have and will i use these two equations to solve for time ? or distance?

im very confused at this point. ( putting two different object's speed together)

2. Nov 26, 2009

Cherry_pie

Re: Kinematics

Hi ya,

The equation you have used is correct - remember, that if the car has caught up with the bus, then d will be the same for both. Hope this helps :)

3. Nov 26, 2009

wisvuze

Re: Kinematics

The question sounds so confusing to me :S If the bus and the car are abreast, then the car has already caught up with the bus..

Anyway, usually for this kind of thing, you have 2 equations, use the variable that is in common with both of them, equate them to each other and solve

4. Nov 27, 2009

charlestchan

Re: Kinematics

d = 15t + 1/2(-0.5)t^2 <<--- for the bus

d = 6t + 1/2(2.5)t^2 <<-- for the sport car.

Solve the simultaneous equation and you had it.

I hope i'm right @_@.