# Rocket car kinematics

## Homework Statement

A "rocket car" is launched along a long straight track att = 0 s. It moves with constant acceleration a1 = 2.9m/s2 .At t = 2.8s , a second car is launched with constant acceleration a2 = 7.0m/s2 .

At what time does the second car catch up with the first one?
How far down the track do they meet?

## Homework Equations

$v_f^2 = v_i^2 + 2a(Δx)$
$v_f = v_i + at$
$x_f = x_i + v_it + 1/2at^2$

## The Attempt at a Solution

$x_f = (1/2)(2.9m/s^2)(2.8s)^2 = 11.368m$

Car 1's X initial is 11.368m when the second car is launched.

Equation for X final of car 1:
$x_f = (0.5)(2.9m/s^2)(t)^2 + (8.12m/s)(t) + 11.368m$

Equation for X final of car 2:
$x_f = (0.5)(7.0m/s^2)(t)^2$

Setting X final equal to each other (when second car meets up with first one)
$(0.5)(7.0m/s^2)(t)^2 = (0.5)(2.9m/s^2)(t)^2 + (8.12m/s)(t) + 11.368m$
$t = 5.06s$

It turns out 5.06s is incorrect. Not sure what I am doing wrong here. Just figured out how to use the fancy math notation! My post is a well-crafted scientific paper.

mfb
Mentor
I can confirm your answer, assuming both cars start from the same place.

Last edited:
jbriggs444
Homework Helper

After 2.8 + 5.06 seconds at 2.9 m/sec^2, where is car number 1?
After 5.06 seconds at 7 m/sec^2, where is car number 2?

Second thing to do is to make sure you are answering the right question. Where does t=0 occur in the problem statement? Where does t=0 occur in your final formula?

I got the answer. I realized that 5.06s is the time after the release of the second car where they meet - not after the release of the first car. I just added 2.8s to 5.06s and got the correct answer: 7.9s.

Thanks for the help!

Then I just plugged values back in to get the answer to the second part, 90m.