# Kinematics ~ need some insight

1. Sep 14, 2005

### SwAnK

hey,

I was given a problem that states:

Alyssa passes by Aaron at a constant velocity of 80km/h. Aaron wanting to catch up to her hesitates for 1 second and then starts accelerating at 10km/h/s. Where does he catch up with alyssa?

OK what i have been thinking and trying out is that the time they meet will be the same right? like when you sub the numbers into an equation the time you get is the same for both? so then you can use that time and figure out the distance, since thats the same too. ALso, so when i get the time thats it right? i dont add the 1 second to it?? the time i got was 14.8 seconds and a distance of 328.9 meters. Did i do this right or am i way off??

2. Sep 15, 2005

### mukundpa

The distance is the same but the time is not same.
Time taken by Aaron in moving the same distance is one second less then taken by Alyssa.
Pay attantion to the units!!!

3. Sep 15, 2005

### moose

D of the constant velocity person is = 80T
the accelerating person's D = .5A(T-1)^2

you can make those equal to each other and solve for T

Last edited: Sep 15, 2005
4. Sep 15, 2005

### HallsofIvy

Staff Emeritus
Uhh, Moose, Distance= 80T

5. Sep 15, 2005

### moose

oops
Changed it, thanks!