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