Game Development Reference
In-Depth Information
Formula for Displacement Given Initial Velocity and Constant Acceleration
∆x = v
0
t + (1/2)at
2
.
(11.15)
Equation (11.15) is one of only a handful of equations in this topic that
are worth memorizing. It is very useful for solving practical problems that
arise in physics simulations.
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It's common that we only need the displacement ∆x, and the absolute
position x(t) doesn't matter. However, since the function x(t) was our
stated goal, we can easily express x(t) in terms of Equation (11.15) by
adding the displacement to our initial position, which we denote as x
0
:
x(t) = x
0
+ ∆x = x
0
+ v
0
t + (1/2)at
2
.
Let's work through some examples to show the types of problems that
can be solved by using Equation (11.15) and its variants. One tempting
scenario is to let our ball bearing hit the ground. The observation deck
on the 103rd floor of Willis Tower is 1,353 ft above the sidewalk. If it
is dropped from that height, how long will it take to fall to the bottom?
Solving Equation (11.15) for t, we have
∆x = v
0
t + (1/2)at
2
0 = (a/2)t
2
+ v
0
t − ∆x
Solving for time
−v
0
±
v
0
− 4(a/2)(−∆x)
2(a/2)
t =
(quadratic formula)
v
0
+ 2a∆x
a
−v
0
±
t =
.
(11.16)
Equation (11.16) is a very useful general equation. Plugging in the numbers
specific to this problem, we have
v
0
+ 2a∆x
a
−v
0
±
t =
(0)
2
+ 2(32 ft/s
2
)(1 353 ft)
32 ft/s
2
−(0) ±
=
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It comes up frequently in job interviews, too.
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