Formula for time to distance from stop

In summary, the time it takes for a vehicle to start from a stopped position on a paved asphalt roadway (coef of approx .70), a grade of +3 degrees, and accelerate to a known distance is dependent upon the force, rate of acceleration, etc. and may not be able to be calculated without many other parameters in place first.
  • #1
jph
3
0
Can someone please post the formula I need to use to determine the time it takes for a vehicle to start from a stopped position on a paved asphalt roadway (coef of approx .70), a grade of +3 degrees, and accelerate to a known distance?

Thanks for any help.

jph
 
Physics news on Phys.org
  • #2
Is this a homework question? In that case, you write what you have done so far.
 
  • #3
As stated this doesn't make a whole lot of sense. Since you give the "coefficient" (of friction?) I suppose we are to calculate the acceleration from force but you haven't told us the force exerted by the engine.
 
  • #4
I think the vehicle is simply rolling downhill. Let the OP present the problem clearly.
 
  • #5
Drive down

This is not for homework or a class assignment. I'm trying to resolve a hunch on how a motor vehicle accident happened. I believe it was intentionally created, called a "drive down."

I am aware of some speed formulas and can derive them from KE=FR.

I believe that there is a formula that will allow me to determine the time it takes for a vehicle to accelerate from a stop to a given distance, and that it is dependent upon the force, rate of acceleration, etc. and I likely may not be able to apply the formula without having many other parameters in place first. I am not sure of what all parameters I must have to derive the time in this case.

I do not recall the necessary elements needed for the formula.

I should be able to obtain them. I know the coef of the roadway, the grade of incline, and the distance the vehicle traveled prior to impact. I know the type of vehicle and can get the weight of the vehicle to an approximation.

Is there a formula, or am I searching for a pipe dream?

Thanks
 
  • #6
Ooh, a mystery thriller!

The grade of just 3 degrees will be negligible, in compared to accelerations produced by cars.

I also presume the co-eff of friction is that for rolling friction. Otherwise, it’s of no use.

Assuming constant acceleration, d = (½)at^2. You can get ‘t’ from this, provided you know ‘a’.

The component of g along the road will be g*(sin 3 deg). The resisting force is 0.7*m*g so the deceleration will be 0.7*g due to friction.

The other component of acceleration has to be experimentally found. You may also get the maximum acceleration from the car manual, but I can’t see how you will know the actual value.

If you can figure out roughly the speed at impact, then you can know the accn by using v^2=2da.
 
  • #7
Thank you!

I appreciate your response. It was very helpful.

jph
 

1. What is the formula for calculating the time it takes to come to a complete stop from a given distance?

The formula for time to distance from stop is: t = √(2d/a) where t is the time in seconds, d is the distance in meters, and a is the deceleration in meters per second squared.

2. How is the formula derived?

The formula is derived from the basic physics equation for motion, which is d = 1/2at² + v₀t + d₀, where d is the distance, a is the acceleration, t is the time, v₀ is the initial velocity, and d₀ is the initial distance. By rearranging the equation to solve for t, we get t = √(2d/a).

3. Can the formula be used for any type of vehicle or object?

Yes, the formula can be used for any object that is decelerating from a given distance. It can be applied to cars, trains, airplanes, or any other moving object.

4. How accurate is the formula?

The formula is a simplified calculation and may not account for all factors such as friction, air resistance, or varying deceleration rates. It can provide a good estimate, but for more precise calculations, additional factors should be considered.

5. Are there any limitations to using this formula?

Yes, the formula assumes a constant deceleration rate and may not accurately reflect real-life situations where the deceleration rate may vary. It also does not take into account external factors such as road conditions, driver reaction time, and braking system efficiency.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
793
  • Introductory Physics Homework Help
Replies
6
Views
724
Replies
207
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
733
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
4K
Back
Top