# Equations of Motion

The following equations only apply to motion in a straight line with constant acceleration. Physics models are often employed to provide these conditions, for example questions often consider air resistance negligible.

Note that you only need to learn the first two equations, and in actual
fact the first equation is the same as acceleration = change in velocity
divided by time taken

**A simple example**

A ball is dropped from a height of 20m with what speed does it hit the
ground.

Acceleration due to gravity = 9.81 ms^{-2} and air resistance is negligible.

The first step is to list what we know and what we want. The question states that he dropped the ball so its initial velocity was zero. We also know the acceleration and the distance and want to find the final velocity.

u = 0

v = ?

s = 20

a = 9.81

t = unknown and not needed

We should therefore choose the equation that omits time.

v2 = u2 + 2as

v2= 0 + 2 x 9.81 x 20

v = 19.8 ms^{-1}

Extended

More complex questions involve motion in a curved path, normally involving some sort of projectile. For these questions the horizontal and vertical components of velocity are separated.

A plot of a ball being thrown including key points.

**Example**

A ball is thrown at 30 degrees to the horizontal with an initial velocity
of 10 ms^{-1} and lands in 0.510 seconds, what is the ball’s
range to 2 s.f. (take acceleration as 9.81 ms^{-2} and air resistance
is negligible)?

The first step should always involve a diagram (though it doesn’t have to be as neat as mine!)

Now we can work of the components of velocity:

Initial Vertical Component of Velocity

10 Sin(30) = 5.00

Initial Horizontal component of Velocity

10 Cos(30) = 8.66

Using horizontal components of velocity it is know that acceleration is zero, we can therefore apply the equations of motion:

t = 0.510

s = ?

a = 0

u = 8.66

s = ut + ½at2

s = (8.66 x 0.510) + (½ x 0 x 0.5102)

s = 4.4 m to 2 s.f.

### Graphs of projectiles

The parachutist

For all of the above examples it has been considered that air resistance is negligible, however as we know this isn’t really the case.

**Before Falling**

Acceleration = 9.81ms^{-2}, the resultant force is zero and the
velocity is zero

**When the object begins to fall**

Air resistance acts upwards, the resultant force equals the weight minus
air resistance and acceleration decreases

**As velocity increases**

The resistive force (air resistance) increases until air resistance equals
weight and therefore the resultant force is zero and acceleration is zero.
This is known as terminal velocity.

**When the parachute opens**

Air resistance increases dramatically so that air resistance is greater
than the force due to weight. The acceleration is negative (the person
decelerates) until they reach a new lower terminal velocity.

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**Module 1**

Useful Physics Resources

Scalars, Vectors & Graphs

Fundamental Equations and Complex Vector Equations

Equations of Motion

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