# Homework Help: One dimensional motion problem =/

1. Aug 28, 2006

### Double D Edd

Hey everyone,

Well, I am sorta stuck on this problem:

Two trains, one traveling at 78 km/h and the other at 135 km/h, are headed toward one another along a straight, level track. When they are 980 m apart, each engineer sees the other's train and applies the brakes. The brakes decelerate each train at the rate of 1.0 m/s^2.

(a) What is the braking distance for the first train?

(b) What is the braking distance for the second train?

(c) Do they both collide? Yes or No.

This is what I did:

For the first part, since the original distance between the trains wasn't 980m, I decided to draw a diagram where they both apply the breaks and with the lease distance between them as 980m. So, I kept the initial position as 980m and tried to figure out the final position (where the first train stops), then took the final velocity as zero. I then applied the [v(t) = at+ v initial] and found t = 21.6s after converting the speed of the first train to m/s. Then I plugged that into the second equation for constant acceleration and ended up with 1213.28m, which I think is ridiculous. What am I doing wrong here? Of course, I can find out the answers for two and three once I get the first part. Any help or hints appreciated.

Edd.

2. Aug 28, 2006

### Galileo

That's correct.

1213.28m is way too much. What formula did you use? You should get something like 233m.

3. Aug 28, 2006

### PPonte

This kinematic equation might be very useful:

$$x = \frac{v_f^2 - v_i^2}{2a}$$

4. Aug 28, 2006

### Double D Edd

I used X(t) = 1/2*a*t^2 + initialV*t + initial position. I put 980 as the initial position. And thanks for clarifying my time part :) And how did you get 233m XD, because that really makes sense, and I am gonna try and get that.

Last edited: Aug 28, 2006
5. Aug 28, 2006

### Farsight

Edd: to get the feel for this sort of thing you need to look up conversion on google. Or do the math. To go from kilometres per hour to metres per seocnd you multiply by .277778. So the first train is going at 21.66 m/s and decelerating at 1m/s and we're cooking on gas. - it's going to take 21+ seconds to stop, during which time it's going 21 metres per second, then 20 metres per second, then 19 metres per second, etc. The average speed during the stopping period is 11 m/s, and it lasts for 21 seconds, so we' re talking about something like 231m to stop. This tells you whether your arithmetic is in the right ball park, and if it is, you can repeat for the other train, and: easy peasy, bob's your knob, sorted!

6. Aug 29, 2006

### Double D Edd

Ohhhh! XD Now I get it. And I think me taking 980m as the initial position was ridiculous.

Thanks a lot, everyone :D.

Edd.