A young woman named Sassy Sue purchases a sports utility vehicle that can accelerate at the rate of

Question

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?

Answer 1

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.

Answer 2

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^2-2t-1)
d=2.44t^2-4.88t+2.44

(3.66/2=1.83)

1.83t^2=2.44t^2-4.88t+2.44
0=0.61t^2-4.88+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 t-1 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)

Answer 3

necropost

Answer 4

Theory: Sassy Sue is Ms. Sue when she was 20ish.