Collision Avoidance - Constant Deceleration

Click For Summary
SUMMARY

The discussion focuses on calculating the necessary deceleration for a train traveling at 161 km/hr to avoid colliding with another train traveling at 29 km/hr, which is 676 meters ahead. The key equations involved are the kinematic equations: v = v0 + at, x - x0 = v0(t) + (1/2)at², and v² = v0² + 2a(x - x0). The solution requires determining the time it takes for the second train to travel while the first train decelerates, ultimately leading to the correct calculation of deceleration needed to stop within the given distance. A common error noted was the incorrect conversion between km/hr and m/s.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Ability to convert units between kilometers per hour and meters per second
  • Basic knowledge of relative velocity concepts
  • Familiarity with solving quadratic equations
NEXT STEPS
  • Study the application of kinematic equations in collision avoidance scenarios
  • Learn unit conversion techniques, specifically between km/hr and m/s
  • Explore relative motion concepts in physics
  • Practice solving problems involving deceleration and distance
USEFUL FOR

Physics students, engineers, and anyone involved in safety analysis for transportation systems, particularly in collision avoidance scenarios.

Beamsbox
Messages
61
Reaction score
0
Alright, once again I've failed myself and now I think I've wasted way too much time on trying to figure this out. I just want a bit of help in setting this thing up and perhaps a short description of which variables I need to find first. Any help, much appreciated of course. Thanks all.

Homework Statement


Basically you have two trains, the first traveling at 161km/hr and the second traveling at 29km/hr and 676m ahead of the first. The first slams on his brakes. What magnitude of deceleration must result if a collision is to be just avoided?


Homework Equations


Given equations, not sure which I need to use first, etc.
v=v0 + at
x-x0 = v0(t)+(1/2)at2
v2 = v02 + 2a(x-x0)
x-x0 = (1/2)(v0+v)t
x-x0 = vt-(1/2)at2

The Attempt at a Solution



So, the various attempts I've made seem to lead me to nowhere. The thing that throws me off the most, is I know that if I find the deceleration across the 676m distance, it won't work because the second train has obviously moved.

I'm thinking that I need to somehow find the amount of time the second train takes to travel the distance that it moves while the deceleration takes place, and then use that to determine the distance it travels. From there I could easily calculate the deceleration over the entire distance. BUT! I don't know how to find that time, and I'm thinking that once you write the CORRECT equations, and substitute them into each other perhaps that time will cancel out... leaving you with all the information you need.

But, like I said, I'm lost. And haven't a clue on where to start. I keep confusing myself from all sorts of different angles.

Thanks again for any help!
 
Physics news on Phys.org
Look at it from the point of view of the leading train. You are at rest and you see a train coming at you with the (relative) velocity of 161 - 29 = 132 km/hr. The distance between you and the train is 676 m. What deceleration does the train need to have to stop over that distance? Use the third kinematic equation that you posted to find out.
 
Finally got the answer I was supposed to get. I had correctly set up the problem (multiple times), but I had apparently converted between km/hr and m/s incorrectly. I stuck with what you were telling me and after a bit I realized my mistake.

Thank you for your time and input, Kuruman.
 

Similar threads

Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K