Game Development Reference
In-Depth Information
To find:
Given these:
Use this:
v 1
a, v 2 , Δs
v 1 =
v 2 2
- 2 aΔs
In cases where acceleration is not constant, but is some function of time, velocity, or
position, you can substitute the function for acceleration into the differential equations
shown earlier to derive new equations for instantaneous velocity and displacement. The
next section considers such a problem.
Nonconstant Acceleration
A common situation that arises in real-world problems is when drag forces act on a
body in motion. Typically, drag forces are proportional to velocity squared. Recalling
the equation of Newton's second law of motion, F = ma , you can deduce that the accel‐
eration induced by these drag forces is also proportional to velocity squared
Later we'll show you some techniques to calculate this sort of drag force, but for now
let the functional form of drag-induced acceleration be:
a = -kv 2
where k is a constant and the negative sign indicates that this acceleration acts in the
direction opposing the body's velocity. Now substituting this formula for acceleration
into the previous equation and then rearranging yields:
a = dv/dt
-kv 2 = dv/dt
-k dt = dv/v 2
If you integrate the right side of this equation from v 1 to v 2 and the left side from 0 to
t , and then solve for v 2 , you'll get this formula for the instantaneous velocity as a function
of the initial velocity and time:
-k ∫ (0 to t) dt = ∫ (v1 to v2) (1/v 2 ) dv
-k t = 1/v 1 - 1/v 2
v 2 = v 1 / (1 + v 1 k t)
If you substitute this equation for v in the relation v = ds/dt and integrate again, you'll
end up with a new equation for displacement as a function of initial velocity and time;
see the following procedure:
v dt = ds; where v = v 1 / (1 + v 1 k t)
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