   The Equations of Motion
Introduction

In earlier units you have been introduced to the concepts of displacement, velocity, and acceleration. Some of these quantities are defined in terms of others, for example,

We can use mathematical equations based on how an object is currently moving to predict its future motion. Mathematical models of the behaviour of moving objects, such as stars and planets, allow predictions to be made about their position and speed at certain times.

The motion of the Earth around the sun is governed by the laws of circular motion but the positions and speeds of objects moving in a straight line can also be modelled mathematically.

Modelling, based on mathematical descriptions of motion and forces, is used extensively to test the viability of engineering structures before construction.

Movement, like that in a roller coaster, usually involves both curved and straight sections. This unit will introduce the mathematical equations governing uniform accelerations or decelerations for objects moving in a straight line.

Some types of linear motion can be represented by straight lines on appropriate graph axes. The gradients of these straight lines is easily determined and gives additional information about the motion. The motion represented in Fig.1 does not produce a straight line graph but the gradient at any specific point can be determined from the tangent.

 Figure 1. Drawing a tangent. The gradient of a distance versus time graph gives the …
• The gradient of a
velocity
An object's velocity states both the speed and direction of motion relative to a fixed reference point.
velocity
versus time graph gives the …
• In mathematical notation any straight line has the equation

y = mx + c

where x and y are the quantities plotted on the x- and y-axes, m is the gradient and c the intercept on the y axis.

The first equation of motion
An object moving with a constant
acceleration
An object's acceleration is its rate of change of velocity.
acceleration
a (sometimes called a uniform acceleration) starts with an initial velocity u and achieves a final velocity v in a time of t seconds.

The quantities u, v, a, and t are linked mathematically by the equation:

Click to make v the subject of the above equation and note that This is called the first equation of motion.

A car starts from rest and accelerates at 2 ms−2. The speed after 7.5 seconds is …
• Find the final speeds after each of the following motions:
•  Acceleration of 3 ms−2 from rest for 4 s 12 m/s 33 m/s 7.5 m/s. Acceleration of 5 ms−2 from 3 ms−1 for 6 seconds 12 m/s 33 m/s 7.5 m/s. Deceleration of 5 ms−2 from 30 m/s for 4.5 seconds 12 m/s 33 m/s 7.5 m/s.
• If an object starts with a speed u and accelerates in the same direction at a ms−2 for t seconds, then it achieves a final speed v ms−1. The motion of this object is represented by the graph shown in Fig.2.

 Figure 2. Graph for a uniform acceleration.  Since the graph shown in Fig.2 is a straight line we can match the equation for the line to the general mathematical straight-line equation y = mx + c To do this we must rearrange v = u + at slightly to show v on the y-axis and t on the x-axis. v = at + u
 By comparing the two equations, we can see that the acceleration corresponds to the gradient of the line, and the intercept on the y-axis indicates the initial speed.

The straight-line nature of the graph in Fig.2 can be explained in terms of the first equation of motion, where the gradient of the line indicates the acceleration and the intercept on the y-axis gives the initial speed.

With the set up shown in Fig.3 the times to accelerate between initial and final speeds can be found.

The time taken to accelerate from 0 to 100 kmh−1 at an acceleration of 5.6 ms−2 is approximately...
• Figure 3. Testing accelerations The second equation of motion
In Fig.2, the fact that the acceleration is uniform between speeds u and v allows us to state that the average speed is   The distance s travelled during t seconds can be calculated using   Substituting our equation for the average speed gives   We can now substitute v = u + at from the first equation of motion to give    This is called the second equation of motion.

The second equation of motion can be expressed as

Match the distances travelled to each of the following motions.
•  Acceleration of 3 ms−2 from rest for 4 s. 108 m 24 m 80 m Acceleration of 5 ms−2 from 3 ms−1 for 6 s 108 m 24 m 80 m Deceleration of 5 ms−2 from 30 ms−1 for 4 s 108 m 24 m 80 m
• To display the second equation of motion in the mathematical form of y = mx + c again requires a slight rearrangement. To plot a graph which will result in a straight line, we must plot distance on the y-axis and time squared on the x-axis. The gradient of the line will then be ½a, so the value of the acceleration can be determined.

Measuring the acceleration due to gravity
Drop the ball in Fig.4 and note how the distance values increase with time as the ball falls.

 Figure 4. Falling ball. The initial
displacement
An object's displacement quotes both its bearing and distances relative to a fixed reference point.
displacement
of the ball from the starting position is 0 m. As the ball falls, its displacement increases. The rate at which the ball falls increases, so the graph is a curved line. The shape of the line shows that the speed is increasing as the ball falls. The ball is accelerating but we cannot as yet determine if the acceleration is constant during the motion.

In the analysis above, we have shown that for a constant acceleration a graph of distance s versus time squared t2 will produce a straight line. Drop the ball in Fig.5 and step through the stages to complete the table and graph.

 Figure 5. Analysing the data. The straight line in the graph of Fig.5 confirms that the ball is falling with a constant acceleration. The gradient of the line is ½a and so the actual acceleration of the ball can be found.

The gradient of the graph in Fig.5 is approximately …
• The gradient of the graph in Fig.5 has units of …
• The magnitude of the acceleration of the falling ball in Fig.5 is approximately …
• A dropped ball accelerates towards the ground because of the gravitational attraction between the earth and the ball. This attractive force acts as an
unbalanced force
Unbalanced forces occur in situations where the force acting in one direction is greater than the force acting in the opposite direction.
unbalanced force
which causes an acceleration. When the ball is relatively close to the earth's surface, the force is constant and causes all objects to accelerate at 9.81 ms−2. This value is often called the acceleration due to gravity g.

The third equation of motion
The second equation of motion states that

and the first states that We can make t the subject of the first equation   Substituting for t into the second equation of motion gives   Simplifying and rearranging gives By combining the first and second equations of motion we have been able to derive the third:

The third equation of motion is particularly useful in calculations where information about time is not given.

A car accelerates uniformly from rest at 5 ms−2. The final velocity after it has travelled 102.4 m is …
• An object is projected upwards with an initial upward velocity of 49 ms−1. How high will rise if the
deceleration
The deceleration of an objects is a measurement of its rate of change of velocity as it slows down.
deceleration
during the upward motion is 9.8 ms−2?
(HINT: When the ball is at its highest point, it is momentarily at rest and v will be 0 ms−1. The height reached will be the distance travelled until the final velocity is 0 ms−1.)
• Testing the third equation of motion
If an object starts with a speed u and accelerates in the same direction at a ms−2 for t seconds, then it achieves a speed v ms−1 and travels a distance s. The third equation of motion links u, v, a, and s and states that v2 = u2 + 2as. The third equation of motion can be rearranged into the mathematical form y = mx + c
 v2 = 2as + u2

For a constant acceleration, a graph of v2 versus s should give a straight line which has a gradient of 2a.

The arrangement shown in Fig.6 uses light gates to determine the speed of a rolling trolley. Place the gate at position A and measure the velocity of the trolley. Reposition the trolley at the top of the runway and move the light gate to position B. Repeat measurements and build up the table of values for the other locations.

 Figure 6. Velocities measured after the trolley has travelled different distances. When the results table is completed, click through the remaining stages to draw a graph of the results. The straight line in the v2 versus s graph confirms that the trolley has a constant acceleration down the slope. The actual acceleration value is half the gradient of the line on the graph.

The gradient of the graph in Fig.6 is approximately …
• The gradient of the graph in Fig.6 has units of …
• The magnitude of the acceleration of the rolling trolley in Fig.6 is approximately …
• Summary

Constant velocities and accelerations can be described with equations which also explain the shape of the lines on motion graphs.

The first equation of motion states that

The second equation of motion states that

The third equation of motion states that

Exercises
1. A car moving in a straight line starts to accelerate uniformly at 4 ms−2. After 7 s it is travelling at a speed of 35 ms−1. Which row in the following table shows the initial speed of the car and its speed 4 s after the beginning of the acceleration?
• Initial speed / ms−1 Speed after 4 seconds / ms−1 A 0 17.5 B 0 16 C 8 24 D 7 23 E 7 16

 Figure 7. 2. In Fig.7 an object is projected vertically upwards with an initial velocity of 44.1 ms−1.

• For what time will it continue to rise? (The deceleration during the upward motion is 9.8 ms−2.)
s   (to 1 d.p.)

What is the velocity of the object 3 s after it is projected vertically upwards?
ms−1 upwards   (to 1 d.p.)

How far is the object above the projection point after 3 s?
m   (to 1 d.p.)

Calculate the maximum height reached by the object?
m   (to 1 d.p.)

Calculate the distance travelled by the object during the 8 s of the motion?
m   (to the nearest whole number)
• 3. An object hits the ground 3.7 s after it is dropped from the top of a high building.

• How far from the top of the building is the object 1 s after it is released?
m   (to 1 d.p.

What is the height of the building?
m   (to 1 d.p.)

How far above the ground is the object 3 s after it is released?
m   (to the nearest whole number)

What is the speed of the object just before it hits the ground?
ms−1 downwards   (to 1 d.p.)
• 4. A car, initially at rest, accelerates uniformly at 5 ms−2.

• Calculate the speed of the car after it has travelled a distance of 40 m.
ms−1   (to the nearest whole number)

Calculate the speed of the car 4 s after the start of the motion.
ms−1   (to the nearest whole number)
• 5. When the car in the previous question is 40 m away from its starting point, the driver stops accelerating. The car then decelerates uniformly to rest after a further 10 s.

• Calculate the total time from when the car starts until it comes to rest.
s   (to the nearest whole number)

Calculate how far the car is away from its starting point after 14 s.
m   (to the nearest whole number)

Calculate the average speed of the car.
ms−1   (to the nearest whole number)
• 6. A sprinter starts from rest and accelerates uniformly over the first 30 m of a 60 m dash. At a point 30 m from the start his speed is 12 ms−1.

• What is his acceleration during the first 30 m?
ms−2   (to 1 d.p.)

How long does it take the sprinter to cover the first 30 m?
s   (to the nearest whole number)

How long does it take the sprinter to complete the final 30 m of the race if he maintains the speed of 12 m−1 right to the finish line?
s   (to 1 d.p.)

The sprinter takes 0.15 s to react to the starting pistol. What is his total time for the 60 m dash?
s   (to 2 d.p.)
• 7. An object starts from rest and moves due east, accelerating at 3 ms−2 for 4 s.

• What is the velocity of this object after 4 s?
ms−1; bearing °   (to the nearest whole number)

After 4 s, how far due east of the starting position is the object?
m   (to the nearest whole number)
• 8. The speed of a lorry 'freewheeling' along a level road reduces constantly from 4 ms−1 to 1.5 ms−1 while the lorry travels a distance of 27.5 m.

• Calculate the deceleration of the lorry?
ms−2   (to 2 d.p.)

How much further will the lorry travel before coming to rest, if the deceleration remains the same?
m   (to 1 d.p.)

What time does the lorry take to come to rest from its initial speed of 4 ms−1?
s   (to the nearest whole number)
• 9. A weekday InterCity express train travels the 300 km between London and York in 3 hours. What is the average speed of the train?
•  kmh−1   (to the nearest whole number)
• 10. The Saturday train completes the same journey in 3 hours but spends 45 minutes stopped at stations. A passenger who catches both the weekday and the Saturday service feels that the Saturday train is travelling faster, yet the train management maintain that the average speed for the journey is the same each day of the week. Which of the following statements is true.
• Figure 8. 11. A dragster in a race, as shown in Fig.8, covers the 440 m from rest in 5.45 s. Calculate the speed of the dragster at the end of the course, if its acceleration is constant throughout the 5.45 s.
•  ms−1   (to the nearest whole number)
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