Game Development Stack Exchange is a question and answer site for professional and independent game developers. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

So I have a a character, say a spaceship. It needs to move distance R, and in direction T (Theta). So say if The object is at (0,0), and it needs to get to (4,3), it has R = 5 and T = about 36 degrees.

Essentially I know where the endpoint is, and the distance away. I want my spaceship to start decelerating when R becomes 8, and the object come to a halt when it reaches the location.

How can I do this. (Kinematics in answers is welcome)

share|improve this question

marked as duplicate by Byte56, msell, bummzack, Anko, Trevor Powell May 1 '13 at 14:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The question has been asked in a variety of forms. Search for arrival questions – Byte56 Apr 30 '13 at 1:58 equations on the wikipedia page as well. – UnderscoreZero Apr 30 '13 at 15:38
@Byte56: I disagree about this being a duplicate of…. That problem concerns specifically how to eliminate oscillation around an intended arrival point, when the approximate equations of motion are used, with variable acceleration. This question asks for how to resolve the equations of motion exactly when constant acceleration is assumed. This answer uses only Grade 11 physics, while the other requires at least 1st year, and perhaps a little 2nd year, university physics. – Pieter Geerkens May 2 '13 at 3:35
I just picked the first arrival question I found. This question also fits. I didn't see constant acceleration implied anywhere in this question, I found the opposite with the "deceleration" wording. – Byte56 May 2 '13 at 11:21
@Byte56: To a physicist, "Kinematics" = constant acceleration; "Dynamics" = (possibility of) non-constant acceleration. Possibly the OP didn't intend that, but it is what the question says; and OP marked my answer as accepted. – Pieter Geerkens May 3 '13 at 22:58
up vote 4 down vote accepted

Here are 3 standard (pre-calculus) kinematics formulae covering the case of constant acceleration, each with one of the unknowns (t, v, or _d) eliminated:

  1. v^2 = u^2 + 2 a d
  2. v = u + a t
  3. d = u t + a (t^2) / 2


  • u and v are the unitial and vinal (sic) velocities respectively;
  • t is the time;
  • d is the distance travelled in time t; and
  • a is the constant acceleration

For your case you know u, v and d, and need to know a, so use formula (1), rearranged into:
a = (v^2 - u^2) / (2 d).

(2) and (3) can then be used to calculate v and d at each time until arrival.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.