Strange case in finding acceleration

AI Thread Summary
The discussion revolves around calculating the acceleration needed for a car to reach a destination at a specified time, using the equation D = V0 t + (1/2)at^2. The user encounters two scenarios: one where the car arrives exactly on time and another where it arrives early, overshoots, and then reverses to reach the destination. The key question is whether there is a constraint on the time variable (t) to ensure the car always arrives on time without overshooting. Suggestions include ensuring the final velocity (Vf) remains non-negative and exploring more accurate numerical integration methods beyond Euler's method to reduce errors. The conversation emphasizes the importance of average speed and numerical stability in achieving the desired outcome.
tungle
Messages
2
Reaction score
0
"Strange" case in finding acceleration

Hello there, looks like a fantastic forum here. I got a problem popped up during my work, hope you can help me out.

I have a car:
- at a distance D away from destination
- current velocity V0
- time to go t : the time period I want the car to spend until reaching destination
I discretize time into seconds. At every every second, I want to find suitable acceleration (or deceleration) so that it can arrive at destination at t.

I apply this equation
D = V0 t + (1/2)at^2

(Given the equation is applied for constant-change in velocity, reapplying it every second yields some error, but it's minor thing)

I came up with something strange:
If t is sufficiently small, then the car arrives exactly at t once. (case 1)

If I set t too high, the car arrives at the destination early (much less than t), go past destination for some time, then go backwards
to reach destination at given t. (case 2)


My question: is there any constraint for t to always yield case 1, preventing case 2 happening.

To be specific, I put an example in the spreadsheet here:
https://spreadsheets.google.com/spr...TGYzRHB1V1RWVjc3ZEE&hl=en_US&authkey=CIyymf8E

Table 2 is where the case 2 happened. Round-up error is ignored.

Thanks much
 
Physics news on Phys.org


D = V0 t + (1/2)at^2

You can solve this equation for a

The final velocity is

Vf = V0 + at

You want Vf >= 0, otherwise the car will pass the end point and then reverse back to it.

Eliminate a and you get an inequality connecting V0, D and t.

You should be able to see what the inequality means physically, in terms of the average speed of the car during the trip.

(Or if you don't like doing math, just think about what the average speed must be to make the final speed >= 0, and check your conclusion with your spreadsheet).
 


Hi,
tks for reply AlephZero.
I know Vf >=0 to make the car stop once. But solving equation
D = V0 t + (1/2)at^2
for a always yields one value, how can you set constraint on a, say if a got a value that violating the constraint, then what else value should it be?
 


You are dealing with numerical error. You can reduce the error by reducing the time step, or by using a different numerical integration method. It looks like you are using Euler Integration, http://en.wikipedia.org/wiki/Euler_method, which is simple but gives bad results. There are more advanced iteration schemes which give more accurate results and are more stable. You might want to try http://en.wikipedia.org/wiki/Leapfrog_integration which is almost as simple as Euler, but much better.
 
Last edited by a moderator:
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?

Similar threads

Back
Top