Interactive textbook: Displacement and velocity

Defining quantities

If you look up distance in a dictionary you will find it has many different meanings. It's the same for displacement. In physics, words mean one thing and one thing only; the meaning is specified clearly in the definition of the word. It is very important that the definitions are remembered so the quantities they represent mean the same thing to everyone.

Remember that physics is about modelling the universe so that we can make predictions (solve problems) the first step is to define quantities.

Defining quantities in context: the night sky

Imagine you are looking at the universe at night. You would see it is made up of lots of bright points - stars and planets.

To model this system you need to look for differences. The most obvious difference is that the stars have different positions. We can quantify these positions by measuring the horizontal and vertical distance compared to some fixed point e.g. the bottom left hand corner. To do this we would need to define a unit, the metre.

As we watch the universe, we would notice some of the points moving.

The position of the point has changed; it has been displaced. The difference between the beginning and end is called the displacement. To get there, this planet has travelled along a curved path. The length of this curved path is defined as the distance.

Distance and displacement

'Distance' and 'displacement' sound very similar. These two quantities even use the same unit (the metre). However, they are quite different.

Distance is used to define how far a body has travelled. This depends upon the path taken so isn't very useful unless we know which way it went.

Displacement defines where the body has ended up relative to its starting position; the journey is irrelevant. To find displacement, we need only the length of the straight line joining the start and finish and its direction. This need for direction makes displacement more than just a number; it is a vector.

Distance, on the other hand, is a scalar.

 

A body moves half way round a circular track of radius 100 m.

What is the distance travelled and displacement of the body?

The circumference = 2πr

 

 

Adding distance and displacement

Consider a body moving in straight lines from A to B to C:

The distance travelled from A to B is \(4\text{ m}\) and the distance from B to C is \(3\text{ m}\). The total distance is \(7\text{ m}\); we simply add the numbers.

The displacement A to B is \(4\text{ m}\) in the \(x\)-direction and the displacement B to C is \(3\text{ m}\) in the \(y\)-direction. The total displacement is \(5\text{ m }\)at an angle of \(37°\) to the \(x\)-axis. To calculate this result we use Pythagoras' theorem and trigonometry.

To add the displacements, we draw them nose to tail. The displacements form a right angle triangle:

Using Pythagoras: \(R^2 = 3^2 + 4^2 = 25\)

\(\Rightarrow R = 5\text{ m}\)

Using trigonometry: \(\tanθ = {3\over 4}\)

\(\Rightarrow θ = 37°\)

 

A body moves 20 m North followed by 20 m West.

What is the magnitude of the final displacement?

√(202 + 202)

 

Instead of using the actual lengths we can use vectors drawn to scale. These are arrows in the same direction as the displacements with length proportional to the magnitude. They form an equivalent triangle that can be used to solve the problem with a ruler and protractor used to measure the overall displacement and angle.

Note that the vectors are not the displacement; they represent the displacement.

Now let's consider a slightly more difficult example where the displacements don't form a triangle on their own. A ship is travelling north while a sailor walks across it.

The ship moves 100 m and the ship is 50 m wide. What is the resultant displacement?

We can draw vectors representing the two displacements. The resultant is found by arranging them nose to tail and calculating the length of the closing side:

Using Pythagoras: \(R^2 = 50^2 + 100^2 = 12500\)

\(\Rightarrow R = 112\text{ m}\)

Using trigonometry: \(\tanθ = {50\over 100}\)

\(\Rightarrow θ = 27°\)

 

sTaking components

You have seen how, using trigonometry, two perpendicular vectors can be added to give one resultant vector. Trigonometry can also be used to split one vector into two perpendicular components.

We can see by scale drawing that the displacement of \(3.6\text{ m}\) at \(56°\) to the \(x\)-axis is equivalent to a \(2.0\text{ m}\) displacement to the East plus a \(3.0\text{ m}\) displacement to the North. Adding the vectors gives a triangle so we can check our answer:

\(2^2 + 3^2=3.6^2\)

Alternatively, applying some simple trigonometry to this triangle we can see that:

\(\cos 56° = {x\over 3.6} \Rightarrow x = 3.6 \cos 56° = 2.0\text{ m}\)

\(\sin 56° = {y\over 3.6} \Rightarrow y = 3.6 \sin 56° = 3.0\text{ m}\)

In general terms, we can use the following relationships:

Vertical component, \(y = R \sin θ\)

Horizontal component,\( x = R \cos θ\)

 

A skier skis 500 m straight down a 30° slope.

What is the change in height of the skier?

The vertical component = 500 sin 30° = 500 x 0.5

Speed and velocity

Thinking back to the night sky (or any other observations of the universe), we can conclude that bodies do not move instantly from one position to another. Sometimes they take a short time and sometimes a long time. So physicists have defined a quantity that relates the distance travelled and the time taken.

Speed is the distance travelled per unit time:

s p e e d space equals space fraction numerator d i s tan c e over denominator t i m e end fraction space space space space space space space space space

If it takes \(2\text{ hours}\) to drive \(100\text{ km}\) between towns A and B, the speed is \(50\text{ km hr}^{-1}\). Assuming that the car speeds up and slows down during the journey, we have actually calculated the average speed. Since both distance and time are scalars, speed is also a scalar.

We can also define a quantity relating the displacement and time. Velocity is the rate of change of displacement:

space space space v e l o c i t y space equals space fraction numerator d i s p l a c e m e n t space over denominator t i m e end fraction

If town B is \(50\text{ km}\) North of town A then the displacement is \(50\text{ km}\) North. The average velocity is then \(25\text{ km hr}^{-1}\) in a Northerly direction. Since displacement is a vector, velocity is also a vector. The direction of the velocity vector is the same as the displacement.

Instantaneous velocity

Often, being able to calculate the average velocity is not very useful. We learn something about the journey as a whole but not what happens at each moment. Imagine that the car returns to town A; the displacement is zero so the average velocity is zero, even though the car has been anything but stationary! 

Instantaneous velocity tells us how fast the car is going at any moment and in which direction it is travelling.

To find an instantaneous velocity, consider a small section of the motion - so small that the line joining the points is straight and the speed has not had time to change:

The magnitude of the velocity is the same as the speed travelled between the points and its direction is in the direction of the line joining the points. This is a tangent to the curve:

As a body moves along a curved path at constant speed (like a car on a roundabout), even though the speed is constant the velocity changes due to the changing direction:

 

A body moves half way round a circular track of radius 100 m in 5 s.

What is the average speed and velocity of the body?

Half the circumference = 100 π so speed = 100 π/5

Displacement = 200 m so velocity = 200/5

 

Since velocity is a vector we must use vector addition. Lets consider the sailor on the boat again:

The time taken for the sailor to cross the boat is \(25\text{ s}\), the velocity of the sailor is \(2\text{ m s}^{-1}\) and the velocity of the boat is \(4\text{ m s}^{-1}\). The resultant velocity can be found by arranging the vectors nose to tail:

  • The resultant velocity, \(v =\sqrt{2^2+4^2} = 4.5\text{ m s}^{-1}\)
  • The direction is \(\tan^{-1}{4\over2} = 63°\) North West

A mouse runs across a moving walkway at 1m s-1.

If the walkway moves at 2 ms-1 what is the magnitude of the mouse's resultant velocity?

Resultant = √(12 + 22)

 

In another general example, we can also take components of velocity:

  • The \(x\)-component of velocity is \(3.16 \cos 18.4° = 3\text{ m s}^{-1} \)
  • The \(y\)-component of velocity is \(3.16 \sin 18.4° = 1\text{ m s}^{-1} \)

Note the direction of the displacement vector tells us in which direction the body has moved. The direction of the instantaneous velocity vector tells us the direction it is moving.

A skier skis 500 m straight down a 30° slope in 10 s.

What is the horizontal component of velocity of the skier?

The vertical component = 50 cos 30° = 50 x 0.866

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