# Trig Physics

1. Feb 22, 2006

### rculley1970

I am having problems with starting a certain problem.

A train traveling at 297 km/h requires 1.45 km to come to an emergency stop. Find the braking acceleration, assuming constant acceleration.

Now I am not given the acceration or time so this one is stumping me. I have tried several formulas including:

deltaX = 1/2(a)(t)^2 + Vo(t)
(v)^2 = (Vo)^2 + 2(a)(delta X)
v = Vo +a(t)

I cannot figure out how to get time or acceleration to solve for the other. The acceleration isn't due to gravity so it isn't (-9.8m/s^2) so I am at a loss for what equation to use. Should I solve for time first? If so, what is the equation I am missing? As far as I know, I am given Vo (297), Vfinal (0), delta Y (-297), delta X (1.45) and I have already tried converting km/h to m/s which the answer is supposed to be in. CONFUSED!!!!!

2. Feb 22, 2006

### nrqed

You wrote the equation that you need!!
$v_f^2 = v_i^2 + 2 a_x \Delta x [/tex]!! That's all you need! 3. Feb 22, 2006 ### rculley1970 I tried that already but will try again. I am using Vf^2 = 0 since that is the final velocity, 297 as the initial velocity, and 1.45 as delta X. I am coming up with -30417 but will keep messing with it to figure it out. I know it CAN'T be this hard. Should I be finding the time for it to stop? 4. Feb 22, 2006 ### jollygood all u need is (v)^2 = (Vo)^2 + 2(a)(delta X) the point here is, to think what happens when u hit brakes. u slow down, which is a deceleration or a negative acceleration. (opposite is speeding up, which is positive acceleration). usually in a problem like this both these situation are referred by the term acceeleration and leave you to decide. gravity does not come in. you're moving horizontally. gravity acts only on objects travelling in vertical direction. so from starting velocity V0=297 km/h to final velocity V=0 (i.e. to a stop) 0 = V0^2 + 2 a (delta X) u know delta X. plug in and solve for 'a', acceleration. final answer will be negative, proving that you are actually decelerating. 5. Feb 22, 2006 ### rculley1970 OK, I have the equation: 0^2 = (82.5)^2 + 2(a)1450 changed km/h to m/s and km to m. I am coming up with -2.35m/s acceleration. If I am wrong let me know. I am busy checking it right now. 6. Feb 22, 2006 ### nrqed right...except for the units.... 7. Feb 22, 2006 ### rculley1970 I know it is probably an easy problem after seeing how it is done but I have been fighting this problem by myself for 4 days and just can't seem to figure out why I can't get the acceleration without the the time. The example in the book shows: A plane brakes at (Ax) 10mi/h, after Vo of 160mi/h. this gives acceleration, initial velocity, final velocity and time can be figured out. 8. Feb 22, 2006 ### rculley1970 Do you mean changing km/h to m/s? 9. Feb 22, 2006 ### rculley1970 Do you mean changing km/h to m/s? 10. Feb 22, 2006 ### nrqed No. You got the right answer, but it is in [itex] m/s^2$, not on m/s

11. Feb 22, 2006

### rculley1970

I am taking a break for the night on it. Email if you can explain the hint a little bit more. Will be at work at it again in the morning. I know it isn't that hard of a problem but I am making it hard. I just need to figure out what I am doing wrong with the conversion for it. Thank you for your help.

12. Feb 22, 2006

### nrqed

? But you are done!! You did find the acceleration!

13. Feb 22, 2006

### nrqed

you got ir right. You converted the distance and speed correctly. I was just pointing out that you gave your final answer with the wrong units. (but it's the correct numerical value)

14. Feb 22, 2006

### rculley1970

you mean -2.35 m/s^2. lol, i thought you meant I had the conversion wrong. Thank you for your help and I am going to redo the problem again in the morning just to verify. Sorry if I didn't get it soon enough but I guess I didn't carry the units across like I should have. I need to work on that. Once again, Thank you for all everyones help. By the way, this homework is already supposed to be submitted but I didn't get it done in time so I got a 0 on it. I am just trying to understand it because it may be on the exam coming up.

15. Feb 22, 2006

### nrqed

Sorry if I made you worry!

And you have the right attitude: it's very important to understand that very well in preparation for the tests.

good luck!