When Do Rocket Cars with Different Accelerations Meet?

[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.

 

[mfb]

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

Edit: Ah, jbriggs found the problem, your answer is not the final answer yet.

 

[jbriggs444]

First thing to do is to double-check your answer.

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?

 

[Ritzycat]

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 into get the answer to the second part, 90m.

 

[HalfLostAndSearching]

Dear student,

Thank you for sharing your solution to the problem of the rocket cars. Your use of kinematic equations is a good approach to solving this problem. However, there are a few errors in your calculations that may have led to the incorrect answer.

Firstly, when finding the initial position of car 1, you have used the formula for displacement (Δx) instead of initial position (x_i). The correct equation to use would be x_i = v_i*t + 1/2*a*t^2, where v_i is the initial velocity and a is the acceleration. In this case, x_i = 0 + 1/2*(2.9m/s^2)*(2.8s)^2 = 11.368m.

Secondly, the equation for the final position of car 1 should be x_f = x_i + v_i*t + 1/2*a*t^2. You have used the incorrect equation x_f = (0.5)*(2.9m/s^2)*(t)^2 + (8.12m/s)*(t) + 11.368m. This should actually be the equation for car 2's final position.

With these corrections, the equation for car 1's final position becomes x_f = 11.368m + (8.12m/s)*(t) + 1/2*(2.9m/s^2)*(t)^2. Similarly, the equation for car 2's final position becomes x_f = 1/2*(7.0m/s^2)*(t)^2.

Setting these equations equal to each other, we get 1/2*(7.0m/s^2)*(t)^2 = 11.368m + (8.12m/s)*(t) + 1/2*(2.9m/s^2)*(t)^2. Solving for t, we get t = 5.06s, which is the correct answer.

To find the distance at which the cars meet, we can substitute this value of t into either of the equations for x_f. Using the equation for car 1, we get x_f = 11.368m + (8.12m/s)*(5.06s) + 1/2*(2.9m/s^2)*(5.06s)^2 = 66.5m. Therefore, the cars will meet