Grade 12 Physics
posted by Haley on .
A young woman named Sassy Sue purchases a sports utility vehicle that can accelerate at the rate of
4.88m/s^2. She decides to test the car by dragging with another speedster, McSpeedy. Both start from rest, but experienced McSpeedy leaves 1.0s before Sue. If McSpeedy moves with a constant acceleration of 3.66m/s^2 & Sue maintains an acceleration of
4.88m/s^2, find the distance she travels before catching him.
So far I have displacement of Sue is 2.44deltat^2
McSpeedy's v2 before Sue starts: 3.66m/s
McSpeedy's displacement: 1.83delta t^2
I am missing parts and can't figure out what
Any ideas?

Her distance is 1/2 4.88 *t^2
His distance is 1/2 3.66 (t+1)^2
set them equal, and solve for t. 
mhm ok so is this what the equation should look like (before I use quadratic equation)?
4.88/2= 2.44
2.44(t+1)^2
= 2.44 (t^22t1)
d=2.44t^24.88t+2.44
(3.66/2=1.83)
1.83t^2=2.44t^24.88t+2.44
0=0.61t^24.88+2.44
I am not even sure how I did the top one before placing it into the 1.83 equation, I am really confused (before, I had t1 instead, and somehow worked out the above equation and then placed that into the 1.83, but now I realize I need to have
t+1, so that makes things a little more difficult)