# Rocket car problem, find final velocity.

1. Apr 11, 2005

### stangeroo

We have to solve this problem, but also do a write-up on it similiar to how the textbook gives example problems. Here is the problem:

Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car. The rocket provides a constant acceleratoin for 9 seconds. As the rocket shuts off, a parachute opens which slows the car at a rate of 5m?s^2. The car passes the judges' box 990 meters from the starting line 12 seconds after you fire the rocket. What is the car's speed as it passes the judges?

So after drawing a pictorial diagram I came up with the following info.

$$V_0$$=0
$$X_0$$=0
$$T_0$$=0

A=?

$$V_1$$=?
$$X_1$$=?
$$T_1$$=9

A=-5

$$V_2$$=?(the main thing youre trying to find)
$$X_2$$=990
$$T_2$$=12

I figure you need to find all the info out about the middle point, when the rocket shuts off and the parachute opens. And it seems like you need to work from the end backword, but I don't know how. Any help would be greatly appreciated

Last edited: Apr 11, 2005
2. Apr 11, 2005

### stunner5000pt

for the first leg (when rockets are fired up)
let the distance (unknown) be d1
t = 9 s
v1 = 0
v2 = ?
$a = a_{rocket}$

find this distance d1 in terms of v1, t and a(rocket)
now what is v2? v2 = v1 + a(rocket) t

now for the second leg of the journey
d = d2 (unknown)
t = 3s
v3 = v2
v4 doesnt matter
a = -0.5m/s^2
relate this distance d2 with the variable and constants that you have i.e. t, v3 = v2, and a.
d1 + d2 = 990
solve for a(rocket)
now solve for v2

3. Apr 11, 2005

### BobG

You know your position equation, right?

$$x_f=x_i+v_i+\frac{1}{2}at^2$$

$$v_f=v_i+at$$

You use both of those equations for both intervals.
Set up your equation for the first interval. Your final position over the first interval is the initial position for the second interval. Since you can't assign a number to the position, yet, you have to substitute the entire equation in as the initial position of your second interval. Once you do that, you'll only have one missing variable left: acceleration over the first interval.

Once you find the acceleration for the first interval, it's easy to find the velocity at the end of the first interval. Substitute that result for the initial velocity of your second interval and solve.

4. Apr 11, 2005

### stangeroo

ok, so the two equations for the first interval is

$$V_f1=9a$$and
$$X_f1=40.5a$$

Now im not sure if i plugged them in right, but I got 20.45 as $$A_1$$

Put 990 as $$X_f$$, 40.5a as $$X_1$$, 9a for $$V_1$$, then -5 as $$a_2$$ and 3 for $$deltaT_2$$

Last edited: Apr 11, 2005
5. Apr 11, 2005

### stangeroo

ok, I got 2169.05 m/s for the velocity as it passes the judges, can someone else do the problem and tell me what you get.

6. Apr 11, 2005

### theCandyman

That is incorrect.

"990 meters from the starting line 12 seconds after you fire the rocket", your answer is after the object slows down. If it ever traveled 2169.05 m/s ending time would be less than a second. I have not done the arithmetic, so the only suggestion I have is to check your units.

Edit: Take note of this, you should always check to see if your answer is logical.

7. Apr 11, 2005

### stangeroo

ok, i think my error came from using the following equation:
$$x_f=x_i+v_i+\frac{1}{2}at^2$$

shouldnt there be $$t$$ after the $$v_i$$

8. Apr 12, 2005

### BobG

Yes. Typo on my part Doesn't your book have these equations :craftily shifts blame: (What's up with the "craftily shifts blame" smiley? - it doesn't work. Oh, I see, wasn't crafty enough.)

9. Apr 12, 2005

### stangeroo

lol, yea I have the equations, but out of laziness I just used the ones in this thread
, thanks for the help though