Posted by **Haley** on Wednesday, September 12, 2007 at 4:36pm.

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?

- Grade 12 Physics -
**bobpursley**, Wednesday, September 12, 2007 at 5:09pm
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.

- Grade 12 Physics -
**Haley**, Wednesday, September 12, 2007 at 5:45pm
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)

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