Game Development Reference
In-Depth Information
Taking the limit as Δ t goes to 0 gives the instantaneous acceleration:
a = lim Δt→0 Δv/Δt
a = dv/dt
Thus, acceleration is the time rate of change in velocity, or, the derivative of velocity
with respect to time.
Multiplying both sides by dt and integrating yields:
dv = a dt
(v1 to v2) dv = ∫ (t1 to t2) a dt
v2 - v1 = Δv = ∫ (t1 to t2) a dt
This relationship provides a means to work back and forth between velocity and accel‐
eration.
Thus, the relationships between displacement, velocity, and acceleration are:
a = dv/dt = d 2 s/dt 2
and:
v dv = a ds
This is the kinematic differential equation of motion (see the sidebar “Second Deriva‐
tives” on page 38 for some helpful background). In the next few sections you'll see some
examples of the application of these equations for some common classes of problems
in kinematics.
Second Derivatives
In Chapter 1 , we explained that you need not worry about the use of derivatives and
integrals in this topic if you're unfamiliar with calculus since we'll show you how to
implement the code to take care of them computationally. That still applies, but here
we've introduced new notation:
d 2 s/dt 2
which is an equation representing acceleration as the second derivative of distance trav‐
eled with respect to time. You can think of second derivatives as just two successive
derivatives in the manner we explained in Chapter 1 . In the case of distance traveled,
velocity, and acceleration, the first derivative of distance traveled with respect to time is