General Uses
These are used to describe the
relationships between the following
kinematic quantities:
○ Distance/displacement
○ Speed/velocity
○ Time
○ Acceleration
When there is an unknown, it can be
solved for when the values of the other
quantities are given
The Four Basic Kinematic
Equations are:
V = V0 + a Δt
V2 = V02 + 2aΔs
S = V0Δt + 0.5 a Δt2
S = (V0 + V)/2 × t
V = V0 + a Δt
E.g. A car starts at rest and accelerates
uniformly at 2 m/s2 for 5 seconds and
stops accelerating from here on.
Calculate its velocity after t = 5 seconds.
Using V = V0 + a Δt, we sub in values 0 for
V0, 2 for a and 5 for t. Solving for V, we
get:
V = 10 m/s
V2 = V02 + 2aΔs
E.g. A train accelerates from 10 m/s to 40
m/s at an acceleration of 1m/s 2. what
distance does it cover during this time.
Using V2 = V02 + 2aΔs, we sub in values
40 for V, 10 for V0 and 1 for a. Re-
arranging to solve for s, we get:
S = 750 m
S = V0Δt + 0.5 a Δt2
E.g. A body starts from rest at a uniform
acceleration of 3 m/s2. how long does it
take to cover a distance of 100m.
Using S = V0Δt + 0.5 a Δt2, we sub in values 3
for a, 0 for V0 and 100 for s. Re-arranging
the equation and solving for t (using the
quadratic formula), we get:
t = 8.51 or -8.51 seconds. As time cannot be
negative, t = 8.51 seconds.
S = (V0 + V)/2 × t
A car decelerates from 20 m/s to 10 m/s
over a period of 10 seconds. How far
does it travel during this time period.
Using S = (V0 + V)/2 × t, we sub in values
20 for V0, 10 for V and 10 for t. Solving
for s, we get:
S = 150m
Note:
All units must be converted such that
they are uniform for different variable
throughout the calculations.
Kinematic quantities that are scalar
CANNOT be negative, hence any such
alternate solutions obtained must be
disregarded.
Standard units for the various
quantities are as follows:
Speed – metres/second
Acceleration – metres/second squared
Distance – metres
Time - seconds