# Rocket car kinematics

• Ritzycat

## 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.

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

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