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