# Acceleration of rocket car

1. Sep 30, 2009

### songoku

1. The problem statement, all variables and given/known data
A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be considered, so the acceleration of the car must not exceed 5g (or five times the acceleration of gravity) during the test. Using the latest materials and technology, the total mass of the car (including the fuel) is 6000 kilograms, and the mass of the fuel is one-third of the total mass of the car (i.e., 2000 killograms). The car is moved to the starting line (and left at rest), at which time the rocket is ignited. The rocket fuel is expelled at a constant speed of 900 meters per second relative to the car, and is burned at a constant rate until used up, which takes only 15 seconds. Ignore all effects of friction in this problem.Find the acceleration a_0 of the car just after the rocket is ignited

2. Relevant equations
kinematic and dynamic

3. The attempt at a solution
Because the fuel is used up in 15 seconds, the rate of the mass used = 2000/15 = 400/3 kgs-1.

That's all I can come up with...

Thanks

2. Sep 30, 2009

### tiny-tim

Hi songoku!
ok … so the change in momentum per second of the fuel is … ?

and so the change in momentum per second of the rocket is … ?

3. Sep 30, 2009

### songoku

Hi tiny-tim :)

The change in momentum per second of the fuel = 400/3 * 900 = 120,000 N

The change in momentum per second of the rocket = The change in momentum per second of the fuel ??

Thanks

4. Oct 1, 2009

### tiny-tim

Hi songoku!
That's exactly right!

ok, now you want the acceleration …

and you have the rate of change of momentum, = d/dt (mv),

so acceleration = … ?

5. Oct 1, 2009

### songoku

Hi tiny-tim

the rate of change of momentum = d/dt (mv)

120,000 = m *d/dt (v) ; m = 6000 kg

a = 20 ms-2

Now another question: find the velocity of the car when the fuel is about to be used up.

I think we consider when t = 15 seconds. And I've found that acceleration at t = 15 s is 30 ms-2 using the same method as you taught me.

But then I don't know how to continue...

Thanks

6. Oct 1, 2009

### tiny-tim

Hi songoku!
Why do you want the acceleration?

Just use conservation of momentum again.

7. Oct 5, 2009

### songoku

Hi tiny-tim

Let : 1 = car ; 2 = rocket

I consider when t = 0 and t = 15

m1u1 + m2u2 = m1v1 + m2v2

0 (because u1 = 0) + 2000*900 = 4000*v1 + 0 (because m2 = 0 when t = 15)

v1 = 450 ms-1???

Thanks

8. Oct 6, 2009

### tiny-tim

Hi songoku!

Yes, that's right …

to put it in words: the momentum gained by the rocket-plus-remaining-fuel equals the momentum lost by the fuel, which is the mass of the fuel times the difference in velocity upon ejection, which at the end is the total mass of fuel times the (constant) relative ejection velocity, ie 2000 times 900.

9. Oct 6, 2009

### songoku

Hi tiny-tim

Thanks a lot for your help !

10. Sep 18, 2010

### cupcake

hi, I have a same question : Find the final velocity of the car just as the rocket is about to use up its fuel supply.

I have tried exactly the same way, I used conservation of momentum, and got the answer Vfinal = 450 m/s.. but, the answer turned out to be wrong.. please advise.

11. Sep 18, 2010

### songoku

Hm...how can you know that the answer is wrong? tiny-tim helped me and I also got the same answer (post #7).

12. Sep 21, 2010

### cupcake

that's why I don't know.. mastering physics said it wrong...
the question is Find the final velocity of the car just as the rocket is about to use up its fuel supply...

I just use the conservation of momentum and got the answer 450 m/s..but, it said the answer is wrong.

13. Sep 22, 2010

### cupcake

14. Sep 25, 2010

### cupcake

up up...

15. Oct 8, 2010

### songoku

What is "mastering physics" ? I think the answer is 450 m/s. Maybe we should wait for other members to give their opinions here.

16. Nov 19, 2010

### Frillo

You should use the rocket equation:

"The differential equation for rocket motion is an example of a separable differential equation. It can be rewritten as

int(dv/dt)*dt = -v_ex * int(1/m)*dm

where the fact that a = dv/dt has been used on the left hand side. Thus, integrating the left side gives the change in velocity. Integrate the right side to get an expression for the change in velocity in terms of the initial and final masses."

This leads to: v_final-v_0 = v_ex*ln(m_0/m_final)

17. Jan 26, 2011

### songoku

how can you get this equation?
and what is -v_ex?

Thanks