# Homework Help: 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