- Describe the principles of a method for measuring acceleration.
- Draw an acceleration–time graph using information obtained from a velocity–time graph for motion with a constant acceleration.
- Use the terms ‘constant velocity’ and ‘constant acceleration’ to describe motion represented in graphical or tabular form.
- Show how the following relationships can be derived from basic definitions in
kinematics: v = u + at, s = ut + ½at
^{2}, v^{2}= u^{2}+ 2as. - Carry out calculations using the above kinematic relationships.
- Equations of motion in horizontal contexts.
- Equations of motion in vertical contexts.
- Equations of motion in projectile contexts.

Aids to Understanding

Measuring Acceleration

Everyone thinks they can describe how to measure acceleration until they try during an exam. Then they wished the had thought about it
more closely. Make sure you don't make that mistake!

Suppose the slope is increased slightly, suggest three different times from the ones in the diagram and calculate a new acceleration
that is consistent with a steeper slope.

Graphs of Motion

The pulses of ultrasound are sent out, reflect of car and received. The time between sending and receiving is recorded and the distance from the transmitter/ receiver to car is calculated. If this distances changes then further calculations deduce velocity and acceleration. The software plots these quantities enabling comparison and analysis. In the above example energy losses are ignored.

Graphs of motion 2

Find the velocity and acceleration time graphs for the following motion.

- A car is travelling at 3.0 ms
^{-1}and accelerates at 2 ms^{-2}for 3 s. - The car travels at a steady speed for 4 s.
- The car gradually comes to rest with an acceleration of -1.5 ms
^{-2} - After waiting 9 s, the car accelerates for 2 s at 0.5 ms
^{-2}. - The car travels at at steady speed for 3 s.
- The car accelerates at -1 ms
^{-2}, until it comes to rest.

Deriving the Equations of Motion

Equation 1 follows from a rearangement of a=(v-u)/t. The authenticity of Equation 4) is self evident ie, the average speed of a body speeding
up from u to v with constant acceleration is (v + u)/2, multiplying by t gives us displacement.

To derive Equation 2, we simply draw a velocity time graph starting from u and speeding up to v in time t. The total area under
the graph is the displacement of the body during that time.

To derive equation 2, we start with v = u + at, then square both sides.

We end up with

v^{2} =u^{2} + 2uat + a^{2}t^{2}

Grouping terms and taking a common factor of 2a

we end up with

v^{2} = u^{2} + 2a(ut + 1/2at^{2})

v^{2} = u^{2} + 2as.

Acceleration is a vector

When treating acceleration as a vector, a negative acceleration does not mean a deceleration and a positive acceleration
does not necessarily mean speeding up. The sign indicates direction, ie the negative (left or down) or positive (right or up) direction.
For example, the force of gravity always pulls down on a body and given the force and the acceleration it causes must always
be in the same direction, the acceleration must also be down, in otherwords, negative.
**Example 1** a ball is travelling upwards with v = +60.0ms^{-1}, after 3 seconds it is now travelling at = +30.6ms^{1}
Use a = (v-u)/t to calculate the ball's acceleration.

**Example 2** a ball is travelling downwards with v = -30.6ms^{-1}, after 3 seconds it is now travelling at = -60.0ms^{1}
Use a = (v-u)/t to calculate the ball's acceleration. Compare your answers.

If the sign of a body's velocity and acceleration are the same, the body gets faster. If the sign of a body's velocity and
acceleration are not the same, the body slows. The example above verifies this.

Click for homework on Equations of motion

Projectile problems

1) Suppose an astronaut on the moon hits a golf ball at the same angle of elevation and speed as the projectile
in the main body text. Complete the exercise and provide a full description of the ball's motion. The Moon's
acceleration is -1.6ms^{-2}.

2) A ball is kicked horizontally of a cliff. If the kicker wishes the ball to land avoid 20 m of water below
the cliff and the cliff is 30 m high. With what minimum speed will the ball need to leave the kicker's boot?

Suppose a toy car rolls down a slope as shown.

Describe this activity and show that the magnitude of car's acceleration is 0.83 ms

acceleration-time graphs. The positive accelerations of the car indicated on the graph are due to the elastic

band.

- A car is travelling at 5.0 ms
^{-1}and accelerates at 1 ms^{-2}for 5 s. - The car travels at a steady speed for 6 s.
- The car gradually comes to rest with an acceleration of -1.0 ms
^{-2} - After waiting 9 s, the car accelerates for 2 sec at -0.5 ms
^{-2}. - The car travels at at steady speed for 1 second.
- The car accelerates at 1 ms
^{-2}, until it comes to rest.

- Stage 1 takes 5 s.
- Stage 2 takes 6s.
- Stage 3 takes 10 s. (Figure out why)
- Stage 4 takes 11s.
- Stage 5 takes 1 s.
- Stage 6 takes 1 s. (Figure out why)

Draw this graph yourself, or try to draw the one suggested in the left hand column. Notice that all accelerations are horizontal lines (constant). The area under each graph also has physical significance. The total area under the v-t graph is the displacement of the car (vt =s) and the area under the a-t graph (in yellow) is equal to the change in velocity (at = v - u). Use this information to verify the velocity of the car after 17.5 s is 3.5 ms

- Equation 1) is for finding the final velocity when we know how long the body has been accelerating for.
- Equation 2) is for find the final velocity when we know the displacement of the body due to its accelaration.
- Equation 3) is for finding its final displacement (Note 1) and 3) have identical terms.
- Equation 4) is simply average velocity x time = displacement

- Calculate the car's final velocity. Answer, use equation 2.
- Calculate the time the car took to travel 10m. Answer use your answer to 1) then use equation 1.

- Runner A starts 1 second before runner B.
- Runner A accelerates at 2.0 ms
^{-2}for two seconds, then travels at a steady speed. - For the final 10 m he decelerates at 0.5 ms
^{-2}until she crosses the line. - Runner B starts 1 second after runner A.
- Runner B accelerates at 2.2 ms
^{-2}for 2.5 s seconds, then runs at a steady speed. - For the final 10 m runner B decelerates at 0.4 ms
^{-2}until he crosses the line.

An answer to this problem has been provided by Katy an S5 pupil. Katy looked at the graph and quickly concluded B won.

Use v

- v
^{2}=u^{2}+ 2as = v^{2}= 25 +2(-9.8x-100) - v = -43.9 ms
^{1} - Now use t = (v-u)/a
- t = (-43.9 - 5)/-9.8
- t = (-48.9)/-9.8 = 4.9s
- The sandbag takes 5s to reach the sea

It can be noted that ...

- The vertical velocity vector of a projectile is subject to the force of gravity and equals -9.8ms
^{-2}throughout the flight. - The horizontal velocity of the projectile remains constant throughout the flight.
- The actual velocity of the projectile is the vector sum of the horizontal and vertical velocities.
- The horizontal and vertical velocities are independent of each other and can be studied separately.

Vertical Information | Horizontal Information |
---|---|

s_{v} = u_{v}t + 1/2a_{v}t^{2} |
s_{h} = u_{h}t (note a_{h} = 0) |

v_{v}^{2} = u_{v}^{2} +2a_{v}s_{v} |
v_{h}^{2} = u_{h}^{2} ie v_{h} = u_{h} |

v_{v} = u_{v} + a_{v}t |
v_{h} = u_{h} |

v_{v} = Vsinq |
v_{h} = Vcosq |

To find the velocity at all points |
---|

v^{2} = v_{v}^{2} + v_{h}^{2} |

Since we know the projectile's vertical acceleration is equal to -9.8 ms

- The initial horizontal and vertical components of velocity of the projectile
- The total time of the flight.
- The maximum height of the projectile.
- The maximum range of the projectile.
- The vertical and horizontal velocities of the projectile at any point during its flight.
- The vertical and horizontal displacements at any point during the flight.
- The velocity of the projectile at any point during its flight.
- The vertical and horizontal velocity-time and acceleration –time graphs for the flight.

- v
_{v}= Vsinq = 60sin60 = 51.6ms^{-1}; v_{h}= Vsinq = 60cos60 = 30ms^{-1} - Use v
_{v}= u_{v}+ a_{v}t. The time for the flight is the time taken for the vertical velocity to change from 51.6ms^{-1}to -51.6ms^{-1}. t = -102.4/-9.8 = 10.4 s^{-1} - The maximum height occurs after 5.2s. Use s
_{v}= u_{v}t + 1/2a_{v}t^{2}; s = 51.6x5.2 + 1/2x-9.8x5.2^{2}= (268.32 - 132.50)m = 135.82 m - Use s
_{h}= u_{h}t = 30x10.2 = 306 m - The vertical velocity at any point during the flight (say 3s) can be worked out using
v
_{v}= u_{v}+ a_{v}t, with t = 3 s. The horizontal velocity is constant at 30 ms^{-1} - See 3 and 4 to find the vertical and horizontal displacements at a given time.
- To find an actual velocity during the flight use
v
^{2}= v_{v}^{2}+ v_{h}^{2} - The graphs are drawn below. Note the horizontal acceleration is zero throughout the flight. Study these graphs closely and make sure you understand how to draw them and how they could be used in a problem solving situation.

Additionally...Are there any other graphs you can think of for the ball on each planet. (Look at the graphs directly above?)