Mechanics Tutorial 6 - Motion in a Straight Line

 
Speed, Velocity, and Acceleration

This topic looks at linear motion, i.e. motion in a straight line.

 

You will be familiar with the simple equations:

 

            speed (m s-1) = distance (m)        

                                    time (s)

 

            velocity (m s-1) = displacement (m)

                                          time (s)

 

In Physics code this is written:

 

We also know acceleration as:

acceleration (m s-2) = change in velocity (m s-1) ÷ time (s)

 

In Physics code:

In both these equations, the strange triangular symbol, D, is Delta, a Greek capital letter 'D', which is the physics code for "change in".

 

The picture below shows the difference between distance and displacement.

 

Suppose we have two towns A and B 10 km apart but either side of a hill.  They are joined by a railway line that is straight, and goes through the hill in a tunnel.  The road goes round the hill and the total journey distance is 25 km.

 

So the distance is 25 km.  The displacement (the straight-line distance in a particular direction) between A and B is 10 km due East.

 

If we go from A to B and back again, the distance is 50 km, but the displacement is 0.

 

Question 1

Why is the displacement 0?

Answer

Question 2

A runner accelerates at a rate of 4 m s-2 to her maximum speed of 9.6 m s-1.  What is the time taken for her to reach this speed?

Answer

 

Information and Communication Technology (Computers) can be used to demonstrate the motion of a vehicle.  Sensors (light gates or ultrasonic detectors) are connected to a computer and the computer will record the data at time intervals.  The computer will plot a graph of the motion.

 

Here are some important definitions.

 

 

Note that some textbooks (and examination boards) use "velocity" as a posh word for speed.  It has a specific meaning, displacement divided by time.

 

Graphical Interpretation of Acceleration

We can represent the movement of objects using a graph, usually plotting time on the x-axis (horizontal) and the speed or distance on the y-axis (vertical). 

 

Displacement (Distance) - Time Graph

Here is a displacement-time graph:

 

For a distance time graph, the gradient would be the speed.

 

Velocity (Speed) - Time Graph

Consider a train accelerating from a station along a straight and level track to a maximum speed and slowing down to a stop at the next station.  The easiest way to show this is with a speed time graph.

 

Acceleration is the gradient of the speed-time graph.

From the graph,

 

Distance is the area under the speed-time graph.  To work out the total distance, we would add the areas of:

 

Question 3

Use the information below in the question.  The train's motion is that described in the graph above.

  • The maximum speed of the train is 25 m s-1

  • The time interval OX is 45 s

  • The time interval XY is 45 s

  • The time interval YC is 20 s         

What is:

a)      The acceleration between O and A

b)      The acceleration between B and C

c)      The distance covered while the train is at constant speed

d)      The total distance.

e)      The average speed?  

Answer

 

The corresponding distance time graph is like this:

 

We can work out the speed at any instant by measuring the gradient of the distance time graph. The curved line tells us that the speed is changing.

 

Acceleration-Time Graph

Acceleration is usually uniform, which means that the speed [velocity] is changing at a constant rate.    However in many real life situations acceleration is not constant.  Therefore the speed [velocity] time graph is not a straight line:

 

Here is an acceleration-time graph for a rocket:

Look at the red line.  When the rocket is full of fuel at t = 0, it will still have a certain value of acceleration.  If it doesn't, it will stay on the ground (and probably explode), which rather defeats the object of having a rocket.  So the graph is NOT proportional, although it is linear.

 

If the acceleration is constant (blue line) it will be a horizontal straight line.  In this case, the blue line represents the acceleration due to gravity, g = -9.8 m s-2.  The minus sign shows that the direction is downwards.

 

If the acceleration is positive, it means that velocity is increasing.  If the acceleration is negative, the velocity is decreasing.  The term deceleration is can be used with speed, but not velocity, for which the correct term is negative acceleration.

 

The area under the graph represents the velocity.  If the area is positive, it means upwards.  If it's negative, it means that the velocity is downwards.

 

The gradient of the acceleration-time graph has no meaning.

 

 

Instantaneous and Average Velocity

It is important to distinguish these two quantities.  Instantaneous velocity is worked out using the gradient of the tangent.

 

The instantaneous velocity gives the velocity at a particular moment in time.  It will not be the same as the average velocity.  The average velocity is the total displacement divided by the total time period.
 

 

Equations of Motion

We can use the Equations of Motion to calculate the speed of an object under different circumstances.  They are sometimes called the kinematics equationsThese are quantities are involved in linear motion, movement in a straight line:

 

Quantity

Physics Code

Units

Distance

s

m

Speed at the start

u

m s-1

Speed at the end

v

m s-1

Acceleration

a

m s-2

Time

t

s

 

1.  Speed at finish = speed at start + change in speed

     change in speed = acceleration × time.

     Speed at end = speed at start + (acceleration × time)

 

2.

 

3.  Distance = average speed × time

 

 

4. 

Maths Window:

Tips for Calculations

1.Write down a formula or equation using standard symbols if possible.

 

                                                                                                s = 4.5 m

2.You may wish to write a "shopping list"                                         u = 0 m

for the quantities in the formula/equation                                           v = not mentioned

but this gets no marks.                                                                       a = ?

                                                                                                 t = 0.90 s

 

 

3.Substitute values into the formula/equation without rearranging first

                                                 4.5 m  = 0  m + 1/2 a × (0.90 s)2                                

(unless you know you won't make daft mistakes).

 

 

4.Rearrange and calculate your final answer.                                      a = 4.5 ÷ 0.405

a = 4.5 m ÷ 0.405 s2

 

 

5.Write the answer using appropriate significant figures and, if necessary, standard form.  Add the proper units.    

 

a =11.1 m s-2 = 11 m s-2 (2 s.f. as data are to 2 s.f.)

 

Check that you have answered what the question actually asked for.

 

If you have done all this underline your answer.

 

Question 4

A car is travelling at 30 m s-1 and takes 10 seconds to accelerate to a new speed of 35 m s-1.  What is its acceleration? 

Answer

Question 5

A brick falls off the top of a wall under construction and drops into a bed of sand 14.5 m below.  It makes a dent in the sand 185 mm deep.  What is:

a)       The speed of the brick just before it hits the sand.

b)       Its deceleration in the sand.

c)       What would happen to a person undergoing that deceleration? 

 

Use g = 9.8 m s-2

Answer

                    

In many examples we can ignore air resistance, although you will know for yourselves that the faster you go on a mountain bike, the harder you have to pedal.  This is because of the effects of friction and air resistance (drag).  We will look at this next in Terminal Velocity.  

 

How are the equations of motion derived graphically?

All the equations of motion are derived from the speed (or velocity) time graph

 

1. The first one is quite easy as it's derived from the area under the graph:

       

 

2.  This graph shows acceleration:

 

 

3. This derivation is a little more complex to understand:

   

 

4. The final derivation requires some manipulation of equations: 

 

How do we select the right equation?

The flow chart below can be used to help you to select the correct equation of motion to use:

 

   

 

Question 6

An aeroplane of mass 5000 kg lands on a runway at a speed of 60 m/s and stops 25 s later.

 

Calculate:

 

a – the deceleration of the aeroplane;

b – the braking force on the aeroplane.

c - the distance taken by the aeroplane to stop.

Answer

 

Using Calculus in Kinematics (Extension)

So far we have used a graphical argument that shows the relationship between displacement, velocity, and acceleration.  In simple graphical treatments of the kind we used above, acceleration is counted as constant.  In reality it is not.  We do not get sudden transitions from acceleration to constant velocity, with sudden graphical inflections.  At university level, a calculus treatment is preferred.  Calculus is a mathematical technique that allows us to work out the gradient of a graph or the area under the graph, if we know the relationship between two quantities. 

 

In kinematics you will see the following:

 

1. Velocity is the rate of change of displacement.  In calculus notation, this is written as:

 This has the general pattern of the formula for velocity.

 

2. Acceleration is the rate of change of velocity.  In calculus notation, this is written as:

  Again this is similar to the formula for acceleration we have seen before:

So what is the difference?  If we are talking about a constant rate of change, there is no difference.  However, if the rate of change is variable, the calculus notation is used for an instantaneous change.  This is summed up in the graph below:

In this graph we see a very irregular increase in the velocity of an object.  We can look at the rate of change in velocity at a particular instant, or we look at the overall change in velocity over a longer period.  The instantaneous change in the velocity is represented by the term dv, while the overall change is represented by Dv

 

We could, of course, take a tangent from the graph at the particular instant and measure the gradient of the tangent.  However it is likely that there will be uncertainty.  With differentiation, there is no uncertainty.

 

Acceleration is related to displacement using:

This is a second derivative, which means a derivative of a derivative.

 

Using Differentiation

Differentiation is about finding the gradient of the graph.  As we have seen above:

 

Maths Window

There are two things that we can do with Calculus:

1. We can differentiate, which means that we find the gradient of the graph of a known relationship.

2. We can integrate, which means that we find the area under the graph of a known relationship.

 

We do these mathematically without having to draw the graph.

 

This is NOT a comprehensive treatment of calculus, but I hope it will help you how to use it in kinematics calculations

 

Differentiation

Differentiation is about determining the gradient of a graph.

 

There are a number of rules of differentiation  We will use only two here. 

  • Added constants differentiate to 0;

  • Powers differentiate according to this formula:

  • Multiplied constants are multiplied with the result of the formula that has been differentiated.  Suppose the constant is b:

 

 

Let us suppose we have a a straight line graph that follows the general relationship:

 

y = mx + c

 If we want to differentiate this, we get:

Therefore:

This tells us that the gradient is m.

 

We can use calculus to work out the velocity at a particular instant.  Here is a graph showing the displacement of an object subject to constant acceleration.  This graph has been drawn using the equation:

The value of the acceleration is 4.0 m s-2.

 

We could, of course, work the gradient of the tangent to give us the velocity at exactly 6.0 s, but there will be uncertainty.  Instead, let's use calculus:

 

Therefore we can substitute:

v = 4.0 m s-2 × 6.0 s = 24 m s-1

 

This is a lot easier that drawing a tangent, then measuring the rise and the run.

 

We can differentiate the equation:

As below:

To give us the familiar:

 

 

 

Using Integration

Integration is about finding the area under the graph.  We have seen how the kinematics equations have been worked out using the area under the velocity time graph.  For some questions we have simply counted the squares under the graph.  The counting of square is both tedious and prone to uncertainty. In simple graphical treatments, acceleration is counted as constant.    In reality it is not. Therefore a mathematical approach is more satisfactory, as it can be quicker and is less prone to uncertainty. 

 

Consider a car accelerating at a rate of a m s-2 over a period of t s.  This is a real world situation, and the acceleration is not constant, but reduces because of the increase in friction from the road and the air resistance.

Suppose we want to find the displacement s between the points A and B.  We know that displacement is the area under the graph.  We could use the equation:

 

The equation works out the area under the orange line in the graph.  The answer it would give would be too low.

 

Alternatively, we could break the area into three little strips as shown in the graph.  So we could work out the area of each little strip and add them together:

 

This will give us a better answer, but it will only be an approximation.  We could make the strips narrower, using a shorter time interval. 

 

Therefore the answer gets closer to the real answer, but it is still an approximation.  However, if we make the time interval infinitesimally small, we end up with the true answer.  This we do by the process of integration.  Instead of writing the width of each strip a Dt, we write dt.  Integration adds up all the little strips to give us the true answer.

 

Maths Window

Integration

Integration is the reverse process of differentiation.  The idea is shown in the picture below:

 

The 2x term is the function.  The function to be integrated is sometimes called the integrand.  Therefore we write this in calculus form as:

 

 

The dx term shows that the little strips go along the x-axis.  The integration symbol is a fancy capital letter 'S', which means "summed together".  So we now write:

 

The C term is a constant.  When we differentiated, the constant that was added to the function had a differentiated value of 0.  Now we are applying the process in reverse, we need to have a definite value for the constant.

 

Here are some rules for integration:

  • A constant is added to the integrated function.  In some cases this might be zero.  In other cases it has a definite value.

  • Constants that multiply a function are multiplied with the result of the integrated function.  Suppose we have a constant, b:

  • The power rule is shown below.  It does not work with x-1.

 

  • The integral of x-1 is shown below:

 

Often you will need to integrate between two points.  You may see an equation like this:

 

This means you have to work out the value of the integral at p and the value of the integral at q, and then subtract one from the other.  This is shown in the picture below.  The constant, C = 0 for the sake of this argument.

 

 

 

Consider an object moving with a constant acceleration, a m s-2 from an initial velocity at t = 0, u  m s-1, to a final velocity, v m s-1 at time t.  This is shown on the graph below:

 

We know that this graph shows the equation:

v = u + at

 

We also know that the area under the graph is the displacement.  So we can can integrate, since we know that:

So we can write:

Therefore:

So let's use that result:

Worked Example

An object has an initial velocity of 6.5 m s-1.  At time t = 0 it accelerates at a rate of change of velocity of 2.85 m s-1 .  What is the displacement between time t = 3.6 s and time t = 8.5 s?

Answer

We can write this in calculus notation:

Therefore: