Adaptive Cruise Control: Maintaining a Safe Following Distance

  • Thread starter Thread starter Dav15
  • Start date Start date
  • Tags Tags
    Control
Click For Summary
Adaptive cruise control in vehicles uses distance measurements to maintain a safe following distance, typically recommended as three seconds. The discussion focuses on how to calculate the necessary speed of the following car to maintain this distance, factoring in acceleration and deceleration. Relevant equations, such as v=u + at and s = s_0 + vt + 1/2 at^2, are mentioned to help frame the problem. The key to solving the question lies in equating the positions of both vehicles at specific times. Overall, the conversation emphasizes the importance of understanding both position and velocity in relation to adaptive cruise control functionality.
Dav15
Messages
2
Reaction score
0
This question is related to adaptive cruise control in new vehicles.
Adaptive cruise control automatically determines the speed at which a vehicle should be traveling by measuring the distance between it, and the car in front (mainly for use on highways)
There is a general rule that three seconds is a safe following distance.
So I suppose that the question is trying to find the acceleration (or deceleration) that the vehicle needs to maintain that gap.
Of course this would cut out when the car reaches a speed, pre determined by the driver (to avoid speeding).Relevant equations

v=u + at

The attempt at a solution

I honestly have no clue, I'm not going to bother with posting my attempt because it is so pitiful xD
 
Last edited:
Physics news on Phys.org


Welcome to PF. No attempt is worthless because it let's people see where you are going wrong so they can help guide you to an answer. Also a lot of people won't help if you don't at least show some effort.

I'm not sure what exactly your question is either.
 


Thanks Bhumble,
My question is, if the distance between the two cars is known, how would I find the speed the following car should be going to maintain a specific distance (that distance is m/s x 3)

I drew a very rough diagram, to try explain what I am asking :)

EDIT: I apologise for my handwriting :)
 

Attachments

  • img042.jpg
    img042.jpg
    39.8 KB · Views: 482


It seems to me that you are interested in the position and with a given distance equivalent to 3 seconds of time.
Are you suppose to be considering acceleration as well?
s = s_0 + \vec{v} t + \frac{1}{2} a t^2

Or are we just interested in position and velocity?

Either way the key to setting up the problem is to equate the positions of the two cars when the car in front is at time = t and the following car is time = t+3.

So
s_a(t=t) = s_b(t=t+3)
Then it's just some algebra to solve for what you are interested in.

The equation that you put as relevant makes me think you are trying to ask a different question than you illustrated though.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 16 ·
Replies
16
Views
13K
Replies
3
Views
1K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K