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)