Kinematic Equations for Catching Runner A

[schang]

okay, this is simple and i feel ridiculous for asking, but I'm reviewing for my final and i do not remember how to do this type of problem..

Runner A runs toward a flag pole at a constant speed of 5.0 m/s. Runner B begins to run toward the flagpole at 3.0 m/s^2 when Runner A is 15 m ahead. When will Runner B catch Runner A?

I know I'll be setting two distance equations equal to each other, but beyond that I don't know what to do.

 

[Hootenanny]

Have you come across any kinematic (suvat) equations?

 

[jachyra]

yeah, from kinematics:

D: distance traveled (m)
A: acceleration m/(s^2)
t: time (s)
D0: initial distance

D = 0.5*A*t^2 + V*t + D0

if you start your stop watch (t=0) when runner B starts to run, then runner A is already 15 m away. therefore:

for runner A:
DA = 5*t + 15

for runner B:
DB = 0.5*3*t^2

where
DA: distance traveled by runner A
DB: distance traveleed by runner B

set DA = DB, you get a quadratic.
solve the quadratic and the right answer is 5.24 s

So it takes runner B be 5.24 s to catch up to runner runner A after he/she starts to run.

If you start your stopwatch when runner A starts to run, then your time is 5.24 + 15/5 = 8.24 seconds

Just double check my math please. hope this helps.

 

[Petkovsky]

Runner B needs to cover some extra distance so are you sure about the equation for runner B?

 

[Feldoh]

Runner B needs to cover some extra distance so are you sure about the equation for runner B?

I think he accounted for it with Runner A ( +15)

 

[Petkovsky]

I think he accounted for it with Runner A ( +15)

Actually yes, I am sorry, I didn't see that.