# In need of help!

1. Sep 9, 2006

### BunDa4Th

A train 450 m long is moving on a straight track with a speed of 82.2 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 17.1 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing. (in second)

I used the formula DeltaX = VoT + 1/2AT^2 where A= 0

450 = 82.2 km/hT --> T = 5.47

From here I am at a lost to if I have done it correctly so far or I been inputing it incorrectly. But I think the step is:

deltaX = 17.1 km/h(5.47)

Again I am not sure if it is correct or not. Please help me out on figuring what I'm doing wrong or if I am at the right track.

2. Sep 9, 2006

### BunDa4Th

okay i figure out i was using the wrong formula instead i should be using v^2=Vo^2 + 2AdeltaX but using that still yeilded me the wrong answer.

Any help would be great.

3. Sep 9, 2006

### mbrmbrg

I noticed that in your first step you set acceleration=0.

Constant acceleration doesn't mean a=0; it means that a=c, where c is some unchanging constant. (Like when something is falls, it undergoes constant acceleration of a=9.81 m/s^2)

You need to solve for a, then use that to solve for t (and remember to convert your units so you don't have seconds and hours [not to mention m and km, which I think you already took care of]).

4. Sep 9, 2006

### BunDa4Th

Yea, I notice I did something wrong and had a bit of help from someone and was able to correct myself and got the right answer.

5. Sep 9, 2006

### mbrmbrg

whoooa, never mind. You could solve for a then plug&play, but as they don't ask you for a, you could just use
$$\Deltax=\frac{1}{2}(v_0+v)t$$
and solve for t, still remembering to make sure all units line up.

Last edited: Sep 9, 2006
6. Sep 9, 2006

### mbrmbrg

growwwl... pardon the formula in the last post, though I hope it's just my computer (as has been known to happen to me).

Delta x = (1/2)(v initial + v final)(t)

7. Sep 9, 2006

### BunDa4Th

It is possible to shoot an arrow at a speed as high as 124m/s.
a) if friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (in M)
b)How long does the arrow stay in the air? (sec)

I just to be pointed in the right directiong because I think I am going in the right direction but I am not sure.

would i set this problem up as 0 = 124 - 9.80T and solve for T which is 12.65.

knowing that I would use DeltaX = V_oT + 1/2AT^2? then to find out how long it stay in the air how would i solve for that.

8. Sep 9, 2006

### BunDa4Th

Thanks a lot mbrmbrg that was a much easier way to solve it and faster. While I took the long way to solve it.

Is it possible for you to look at my second problem?

9. Sep 9, 2006

### mbrmbrg

I'd do part a) first, using
to solve for X (keeping in mind that when the arrow reaches its highest point, its velocity is zero).
Then use x in a different equation to find time. But remember: depending on how you set this up, you might get time for half of its flight only (so either set up two equations, or, more simply, in this symmetric case just multiply by two).

10. Sep 10, 2006

### BunDa4Th

Thanks for the help. I finally got the answer.