Help with motion in one direction

  • Thread starter Thread starter saber1357
  • Start date Start date
  • Tags Tags
    Direction Motion
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving motion in one direction, specifically focusing on two cars where one is trailing the other. The scenario includes variables such as speed, braking distance, and time intervals during which the trailing car's attention is diverted.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the equations of motion and question how to set up the problem, particularly regarding the timing of braking for both cars. There are discussions about defining piecewise functions for the positions of each car based on their respective accelerations and the timing of events.

Discussion Status

Participants are actively engaging with the problem, raising questions about the interpretation of the braking times and the initial conditions. Some have suggested breaking down the problem into segments based on time, while others are clarifying the initial setup and assumptions. There is no explicit consensus yet, but the dialogue is productive.

Contextual Notes

There are uncertainties regarding the timing of the braking for both cars, with some participants questioning the wording of the problem statement. This has led to discussions about how to accurately model the motion based on the given information.

saber1357
Messages
67
Reaction score
1

Homework Statement


You are trailing a car by 25m; both you and the car are traveling at 110km/h. Your attention is diverted for 2.0s. At the beginning of that 2.0s, the car in front of you brakes at 5.0 m/s^2.
Suppose your attention is diverted another .40s. If you too break at 5.0m/s^2, what is your speed when you hit the car?

Homework Equations


v^2 = u^2 + 2at ?
x = x0+(v0*t) + 1/2at^2 ?

The Attempt at a Solution


I converted everything to m/s. However, I do not understand which equation I can use that incorporates velocity, time, and position. I'm assuming that I'm suppose to create functions for both cars and set them equal to each other.
Any help will be greatly appreciated.
 
Physics news on Phys.org
The form of the position equations will change everytime someone changes acceleration. So you need to define the functions piecewise. What do the equations look like for t<2.4 sec? How about for t>2.4 sec?
 
Well, if I am understanding you correctly, then the function for the second car is
2ndcar position = 25 + (30.6m/s)*(2.0s) + .5*(-5m/s^2)*(2.0s) when t<2.4
It doesn't really make sense to me. The second car is already breaking at t=2.0s, while I begin breaking at t=2.4s. Unless the equation for my car is position = (30.6m/s)*(2.4) + .5*(-5m/s^2)*(2.4s) when t>2.4. But then what can I do with them to figure out the velocity. I am pretty sure I can figure out the velocity if I somehow solve for the time it takes for me to crash into the second car. But again, I am not sure how to solve for time in this situation.
 
As I read the question the second car was already braking at t=0. Or am I wrong? Can we sort this out tomorrow? You don't have an unknown t in either equation and you should. If you aren't tired, maybe somebody else can pick this up. I'm BEAT.
 
:)
I'm pretty sure this problem states that the second car is breaking at the beginning of the 2.0s, meaning it's breaking at t=2.0s.
Looking forward for the help tomorrow. Thanks for your effort <3
I'm definitely going to try to solve this thing.
 
Maybe you'll get it before I do. Good night!
 
Try approaching it this way. First figure out where both cars are at t=2.4 sec and what their velocity and acceleration are. Now pretend that's a whole new problem. My question was whether the second car was braking starting at t=0?
 
Now that I look again at the equation, perhaps you are right, which means that I'm braking at t=2.4..
 
I have questions about the problem statement since nothing seems to really happen at 2 sec if the other car was already braking "at the beginning of that 2 sec". Sure it shouldn't read "at the end of that 2 sec"?
 
  • #10
This is indeed a confusing question, I'm going to take it up with my professor.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K