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^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)

necropost

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

To solve this problem, we will use the equations of motion to find the distance traveled by Sassy Sue before she catches up to McSpeedy.

Let's start by finding the time it takes for Sue to catch up to McSpeedy. We know that Sue starts 1.0s after McSpeedy, so we can represent the time as delta t - 1.0s.

Next, let's find the displacement of Sue and McSpeedy. We have the equation for Sue's displacement as: displacement_sue = 1/2 * acceleration_sue * (delta t - 1.0s)^2.

Similarly, we have the equation for McSpeedy's displacement as: displacement_mcspeedy = 1/2 * acceleration_mcspeedy * (delta t)^2.

To find the distance traveled by Sue before catching up to McSpeedy, we need to find the difference in their displacements. So, the distance traveled by Sue is:

distance_sue = displacement_sue - displacement_mcspeedy.

Plugging in the given values for the accelerations of Sue and McSpeedy (4.88m/s^2 and 3.66m/s^2 respectively), we can proceed with the calculations:

displacement_sue = 1/2 * 4.88m/s^2 * (delta t - 1.0s)^2

displacement_mcspeedy = 1/2 * 3.66m/s^2 * (delta t)^2

distance_sue = displacement_sue - displacement_mcspeedy

Now, you can substitute the given values into the above equations and simplify to find the distance traveled by Sue before catching up to McSpeedy.

Please note that the missing values (such as the specific time or any additional information) may be required to solve the problem accurately.