Game Development Reference
InDepth Information
You can derive a similar formula for displacement as a function of velocity, acceleration,
and time by integrating the differential equation:
v dt = ds
with the formula derived earlier for instantaneous velocity:
v
2
= v
1
+ at
substituted for
v
. Doing so yields the formula:
s
2
= s
1
+ v
1
t + (a t
2
) / 2
In summary, the three preceding kinematic equations derived are:
v
2
= v
1
+ a t
2
v
2
2
= 2a (s
2
− s
1
) + v
1
2
s
2
= s
1
+ v
1
t + (a t
2
) / 2
Remember, these equations are valid only when acceleration is constant. Note that ac‐
celeration can be 0 or even negative in cases where the body is decelerating.
You can rearrange these equations by algebraically solving for different variables, and
you can also derive other handy equations using the approach that we just demonstrated.
For your convenience, we've provided some other useful kinematic equations for con‐
stant acceleration problems in
Table 21
.
Table 21. Constant acceleration kinematic formulas
To find:
Given these:
Use this:
a
Δt, v
1
, v
2
a = (v
2
 v
1
) / Δt
a
Δt, v
1
, Δs
a = (2 Δs  2 v
1
Δt) / (Δt)
2
a
v
1
, v
2
, Δs
a = (v
2
2
 v
1
2
) / (2 Δs)
Δs
a, v
1
, v
2
Δs = (v
2
2
 v
1
2
) / (2a)
Δs
Δt, v
1
, v
2
Δs = (Δt / 2) (v
1
+ v
2
)
Δt
a, v
1
, v
2
Δt = (v
2
 v
1
) / a
Δt
a, v
1
, Δs
Δt =
(
v
1
2
+ 2
aΔs

v
1
) /
a
Δt
v
1
, v
2
, Δs
Δt =(2Δs) / (v
1
+ v
2
)
v
1
Δt, a, v
2
v
1
= v
2
 aΔt
v
1
Δt, a, Δs
v
1
= Δs/Δt  (aΔt)/2