1.1 Equations of Motion – Physics@Hyndland

Vectors

Vectors and Scalars (Revision of National 5)

It is possible to split up quantities in physics into two distinct groups, those that need a direction and those that don’t. Some are obvious – it makes sense that force has direction; you can push or pull but you need to specify the direction.

It would be nonsense to give a direction to time. To say: “It took 5 seconds East” just isn’t right. It is important that you are familiar with which quantity falls into which grouping.

A scalar is a quantity that can be described by just a size and a unit. e.g. time – 30 s, mass – 20 kg.

A vector is a quantity that is fully described with a size and direction.

e.g. force – 50 N downwards velocity – 20 ms^-1 East.

Adding Vectors (Revision of National 5)

This is more difficult than adding scalars as the direction of the vectors must be taken into account.

The addition of two vectors is called the resultant vector.

When you add vectors they have to be added tip-to-tail.

What does this mean?

Each vector must be represented by a straight line of suitable scale.

The straight line must have an arrow head to show its direction. i.e.

tail tip

The vectors must be joined one at a time so that the tip of the previous vector touches the tail of the next vector. i.e.

A straight line is drawn from the starting point to the finishing point and the starting angle is marked.

The resultant should have 2 arrow heads to make it easy to recognise.

If using a scale diagram the length and direction of this straight line gives the resultant vector.

Alternatively you can use trigonometry and SOHCAHTOA or the sine or cosine rule to calculate the resultant.

Distance and Displacement (Revision of National 5)

The distance travelled by an object is the sum of the distances of each stage of the journey.

Since each stage has a different direction, the total distance has no single direction and therefore distance is a scalar.

The displacement of an object is the shortest route between the start and finish point measured in a straight line.

Displacement has a direction and is a vector.

Consider the journey below. A person walks along a path (solid line) from start to end.

They will have walked further following the path than if they had been able to walk directly from start to end in a straight line (dashed line).

The solid line denotes the distance = 3km. The dashed line denotes the displacement = 2.7 km East

Example

A woman walks her dog 3 km due North (000) and then 4 km (030).

Find her

a) distance travelled
b) displacement.

Solution – Use a ruler to measure the lengths of the vectors and a protractor to measure the bearing.

Choose an appropriate scale e.g. 1cm : 1km
Mark the start point with an X, draw a North line and draw the first vector.
Draw a North line at the tip of this vector and now draw the second vector (tip to tail)
Draw the resultant vector from start to end using the double arrow.
Measure the length of the line and the bearing.
When measuring bearings remember – from START – CLOCKWISE – from NORTH

Speed and Velocity (Revision of National 5)

Speed is defined as the distance travelled per second and is measured in metres per second, or ms^-1. Since distance and time are both scalar quantities then speed is also a scalar quantity.

From previous work in Maths and Physics we know that speed is calculated from the equation:

v = d / t

The velocity of an object is defined as the displacement travelled per second.

Since displacement is a vector quantity that means that velocity is also a vector and has the symbol v.

The equation for velocity is:

v = s / t

Example

A runner sprints 100 m East along a straight track in 12 s and then takes a further 13 s to jog 20 m back towards the starting point.

(a) What distance does she run during the 25 s?

(b) What is her displacement from her starting point after the 25 s?

(d)What is her velocity?

Solution – always draw the vector diagram.

a) d = 100 + 20 (b) s = 100 + ( − 20 )

d = 120 m s = 80 m (090)

c) v = d / t (d) v = s / t

v = 120/25 v = 80/25

v = 4.8 ms^-1 v = 3.2 ms^-1 (090)

Resolving Vectors

We have seen that two vectors can be added to give the resultant using vector addition.

Can we split a resultant vector into the two individual vectors that make it up?

Consider the following.

This shows a resultant vector, V, at an angle Ө to the horizontal.

To travel to the end of the vector we could move in a straight line in the X direction and then a straight line in the Y direction as shown below.

But how do we find out the size of each line?

Since we have a right angled triangle with a known angle we can name the sides.

This means we can use Pythagoras’ theorem to work out the unknown sides.

horizontal component vertical component

V_H = VcosӨ V_V = VsinӨ

Example

A football is kicked at an angle of 70^o at 15 ms^-1.

Calculate:

Solution

a) the horizontal component of the velocity; V_H = VcosӨ = 15cos70 = 5.2 ms^-1
b) the vertical component of the velocity. V_v = VsinӨ = 15sin70 = 14.1 ms^-1

The 3 Equations of Motion

The 3 Equations of Motion (s u v a t)

The equations of motion can be applied to any object moving with constant acceleration in a straight line. You must be able to:

select the correct formula;
identify the symbols and units used;
carry out calculations to solve problems of real life motion; and
carry out experiments to verify the equations of motion.

You should develop an understanding of how the graphs of motion can be used to derive the equations. This is an important part of demonstrating that you understand the principles of describing motion, and the link between describing it graphically and mathematically.

Equation of Motion 1 : v = u + at

a= v-u

at = v – u

u + at = v

v = u + at

Example

A racing car starts from rest and accelerates uniformly in a straight

line at 12 ms^-2 for 5.0 s. Calculate the final velocity of the car.

Solution LIST s u v a t

u = 0 ms^-1 (rest)

v v = u + at

a = 12 ms^-2 v = 0 + (12 x5.0)

t = 5.0 s v = 0 + 60

v = 60 ms^-1

Equation of Motion 2: s = ut + ½ at²

The displacement, s, is the area under the graph:

Area 1 = ut

Area 2 = ½ (v-u) t

But from equation 1 we get that (v-u) = at

So Area 2 = ½ (at)t

Therefore,

s = ut + ½ (at)t

s = ut + ½ at²

Example

A speedboat travels 400 m in a straight line when it accelerates uniformly

from 2.5 ms^-1 in 10 s. Calculate the acceleration of the speedboat.

Solution

s = 400 m s = ut+ ½ at²

u = 2.5 ms^-1 400 = (2.5 x 10) + (0.5 x a x10²)

v 400 = 25 + 50a

a = ? 50a = 400 – 25 = 375

t = 10 s a = 375/50

a = 7.5 ms^-2

Equation of Motion 3: v² = u² + 2as

We have already found that

v = u + at

v² = (u + at)²(square both sides)

v² = u² + 2uat + a²t²

v² = u² + 2a(ut + ½ at²)

And since s = ut + ½ at²

v² = u² + 2as

Example

A rocket is travelling through outer space with uniform velocity. It then accelerates at 2.5 ms^-2 in a straight line in the original direction, reaching 100 ms^-1 after travelling 1875 m.

Calculate the rocket’s initial velocity.

Solution

s = 1 875 m v²= u² + 2as

u = ? 100² = u² + (2 x 2.5 x 1875)

v = 100 ms^-1 10 000 = u2 +9375

a = 2.5 ms^-2 u² = 10 000 – 9375

t u = 25ms^-1

The 3 Equations of Motion with Decelerating Objects

When an object decelerates its velocity decreases. If the vector quantities in the equations of motion are positive, we represent the decreasing velocity by use of a negative sign in front of the acceleration value.

Example 1

A car, travelling in a straight line, decelerates uniformly at 2.0 ms^-2 from 25 ms^-1 for 3.0 s.

Calculate the car’s velocity after the 3.0 s.

Solution

u = 25 ms^-1 v = u + at

v = ? v = 25 + (-2.0 x 3.0)

a = -2.0 ms^-2 v = 25 + (-6.0)

t = 3.0 s v = 19 ms^-1

Example 2

A greyhound is running at 6.0 ms^-1. It decelerates uniformly in a straight line at 0.5 ms^-2 for 4.0 s.

Calculate the displacement of the greyhound while it was decelerating.

Solution

s = ? s = ut + ½ at²

u = 6.0 ms^-1s = (6.0 x 4.0) + (0.5 x -0.5 x 4.0²)

v s = 24 + (-4.0)

a = -0.5 ms^-2 s = 20 m

t = 4.0 s

Example 3

A curling stone leaves a player’s hand at 5.0 ms^-1and decelerates uniformly at 0.75ms^-2 in a straight line for 16.5 m until it strikes another stationary stone.

Calculate the velocity of the decelerating curling stone at the instant it strikes the stationary one.

Solution

s = 16.5 m v² = u² + 2as

u = 5.0 ms^-1v²= 5.0²+ (2 x -0.75 x 16.5)

v = ? v²= 25 + (-24.75)

a = -0.75 ms^-2 v = √0.25

t v = 0.5 ms^-1

Graphing Motion

Graphs

In all areas of science, graphs are used to display information.

Graphs are an excellent way of giving information, especially to show relationships between quantities.

In this section we will be examining three types of motion-time graphs.

Displacement-time graphs

Velocity-time graphs

Acceleration-time graphs

If you have an example of one of these types of graph then it is possible to draw a corresponding graph for the other two factors.

Displacement – time graphs

This graph represents how far an object is from its starting point at some known time. Because displacement is a vector it can have positive and negative values. (+ve and –ve will be opposite directions from the starting point).

OA – the object is moving away from the starting point. It is moving a constant displacement each second. This is shown by the constant gradient. What does this mean?

gradient = = velocity

We can determine the velocity from the gradient of a displacement time graph.

AB – the object has a constant displacement so is not changing its position, therefore it must be at rest. The gradient in this case is zero, which means the object has a velocity of zero [at rest]

BC – the object is now moving back towards the starting point, reaching it at time x. It then continues to move away from the start, but in the opposite direction. The gradient of the line is negative, indicating the change in direction of motion.

Converting Displacement – time Graphs to Velocity-time Graphs

The velocity time graph is essentially a graph of the gradient of the displacement time graph. It is important to take care to determine whether the gradient is positive or negative.

The gradient gives us the information to determine the direction an object is moving.

There are no numerical values given on the graphs above. Numbers are not needed to allow a description. The will need to be used however if we were to attempt a quantitive analysis.

Velocity – time Graphs

It is possible to produce a velocity time graph to describe the motion of an object. All velocity time graphs that you encounter in this course will be of objects that have constant acceleration.

Scenario: The Bouncing Ball

Lydia fires a ball vertically into the air from the ground. The ball reaches its maximum height, falls, bounces and then rises to a new, lower, maximum height.

What will the velocity time graph for this motion look like?

First decision: The original direction of motion is up so upwards is the positive direction

Part One of Graph

Now we need to think, what is happening to the velocity?

The ball will be slowing down whilst it is moving upwards, having a velocity of zero when it reaches maximum height. The acceleration of the ball will be constant if we ignore air resistance.

Part Two of Graph

Once the ball reaches its maximum height it will begin to fall downwards. It will accelerate at the same rate as when it was going up. The velocity of the ball just before it hits the ground will be the same magnitude as its initial velocity upwards

Part Three of Graph

The ball has now hit the ground. At this point it will rebound and begin its movement upwards.

In reality there will be a finite time of contact with the ground when the ball compresses and regains its shape. In this interpretation we will regard this time of contact as zero. This will result in a disjointed graph.

The acceleration of the ball after rebounding will be the same as the initial acceleration. The two lines will be parallel.

This now is the velocity time graph of the motion described in the original description.

Converting Velocity – time Graphs to Acceleration – time Graphs

What is important in this conversion is to consider the gradient of the velocity-time graph line.

In our example the gradient of the line is constant and has a negative value. This means for the entire time sampled the acceleration will have a single negative value.

All acceleration time graphs you are asked to draw will consist of horizontal lines, either above, below or on the time axis.

Reminder from National 5

The area under a speed time graph is equal to the distance travelled by the object that makes the speed time graph.

In this course we are dealing with vectors so the statement above has to be changed to:

The area under a velocity time graph is equal to the displacement of the object that makes the speed time graph.

Any calculated areas that are below the time axis represent negative displacements.