…develop our understanding of using numbers and equations to describe motion.

What do we understand about 'acceleration'? Acceleration is the rate of change of velocity An acceleration of 2 ms -2 means that every second the velocity increases by 2ms -1 An acceleration of -2 ms -2 means that every second the velocity decreases by 2ms -1

Rearrange this to give v =… v = u + at This is one of the Equations of Motion

when acceleration is constant (uniform) and motion is in a straight line

You need to be able to:  select the correct formula  identify symbols and units  carry out calculations to solve the problems of real-life motion.

You need to be able to:  carry out experiments to verify the equations of motion

To do this fully, you might find it an interesting challenge to…  understand where the equations come from.

Label the formula using correct symbols and units  v = u + at

Describe the motion of this object

How can we determine the displacement of the object?

Area under the graph = = 6000 m Area 1 = 5 × 300 = 1500 m Area 2 = ½bh = ½ × (35 –- 5) × 300 = 4500 m

Area 1 = ut Area 2 = ½bh = ½ × (v – u) × t Since v = u + at so v – u = at vu0vu0 t

Area 1 = ut Area 2 = ½bh = ½ × (v – u) × t = ½ × at × t = ½ × at 2 vuvu t Area under the graph = displacement s

Area 1 = ut Area 2 = ½bh = ½ × (v – u) × t = ½ × at × t = ½ × at 2 vuvu t s = ut + ½at 2

 v = u + at Start with Equation 1 and square it  v 2 = (u + at) 2  v 2 = u 2 + 2uat + a 2 t 2  v 2 = u 2 + 2a(ut+ ½at 2 )  v 2 = u 2 + 2as As s = ut + ½at 2 equation 2

Equations of motion Third year v = d ÷ t no acceleration a = (v – u) ÷ t Nat 5 a = (v – u) ÷ t uniform acceleration distance = area under speed time graph displacement = area under velocity time graph Advanced Higher - accelerations which are not uniform Higher v = u + at uniform acceleration s = ut + ½ at 2 v 2 = u 2 + 2as v = ½( u + v) - very fast speeds, relativity

Equations of Motion Aim: To verify that the equation s = ut +½at 2 is true. Labelled diagram listing all apparatus. Measurements Left hand side Use metre stick to measure length of slope, = Right hand side Trolley starts from rest so initial velocity, u = 0 Use QED and light gate to find the acceleration, a, down the slope. Use a stop watch to time how long the trolley took to travel down the slope. Calculations: ut +½at 2 Does this equal distance down the slope? Is it close? Why might it not be exactly equal? If time try to verify another equation of motion.

What do we need to think about when using the equations of motion?

 When dealing with vector quantities we  must have both magnitude and  direction.  When dealing with one-dimensional  kinematics (ie motion in straight lines) we  use + and – to indicate travel in opposite  directions.

 Normally we use the following convention positive + negative – positive + negative – Take care – in some questions the sign convention is reversed

 v = u + at What does a positive value of acceleration mean? –ve +ve

Christine Arron is a 100m sprint athlete. Immediately the starting pistol is fired, Christine accelerates uniformly from rest, reaching maximum velocity at the 50m mark in 4.16 s. Her maximum velocity is m s –1. Calculate her acceleration over the first 50 m of the race, showing full working. –ve+ve

Her acceleration is 2.52 m s –2. In this case, acceleration is a rate of change of velocity with time, with which we are familiar. A positive value means, in this case, increasing velocity with time. What else might it mean? –ve+ve

As she passes the finish line, Christine begins to slow down. She comes to rest in 8.20 s from a velocity of 9.73 m s –1. Calculate her acceleration, showing full working. –ve+ve

Her acceleration is a = –1.19 m s –2. Notice that the acceleration has a negative value. Explain this. –ve+ve

Now consider Christine running in the opposite direction. Notice that the sign convention remains the same. –ve+ve

Immediately the starting pistol is fired, Christine accelerates uniformly from rest, reaching maximum velocity at the 50m mark in 4.16 s. Her maximum velocity is –10.49 m s –1 (why is it negative?). Calculate her acceleration over the first 50 m of the race, showing full working. –ve+ve

Her acceleration is –2.52 m s –2. What does the negative mean? –ve+ve

As she passes the finish line, Christine begins to slow down. She comes to rest in 8.20 s from a velocity of –9.73 m s –1. Calculate her acceleration, showing full working. –ve+ve

Her acceleration is a = 1.19 m s –2. Notice that the acceleration has a negative value. Explain this. –ve+ve

A positive value means gaining speed while moving in the positive direction. OR A positive value means the object is losing speed while moving in the negative direction.

A negative value means the object is gaining speed while moving in the negative direction. OR A negative value means the object is losing speed while moving in the positive direction. Tricky – be very careful with the signs when using equations of motion.

Step 1: Write down the sign convention. Step 2: Write down what you know (think suvat). sdisplacement uinitial velocity vfinal velocity aacceleration ttime Step 3: Any other information, eg acceleration due to force of gravity?

Step 4: Select formula – use data sheet. Step 5: Substitute values then rearrange formula. Step 6: Write the answer clearly, including magnitude and direction, and units.

Usain Bolt is a Jamaican sprinter and a three-times Olympic gold medallist. Immediately the starting pistol is fired, Usain accelerates uniformly from rest. He reaches 8.70 m s -1 in 1.75 s. Calculate his displacement in this time. –ve+ve

s = ? m u = 0 m s –1 v = 8.70 m s –1 a = ? t = 1.75 s –ve+ve

Step 1: Write down the sign convention. Step 2: Write down what you know (think suvat). s = u = v = a = t = Step 3: Any other information, eg acceleration due to force of gravity Now try tutorial questions on Equations of Motion pages 3 to 5 Purple books Ex 1.3 Step 4: Select formula – use data sheet. Step 5: Substitute values then rearrange formula. Step 6: Write the answer clearly including magnitude, direction, and units.

 …our understanding of using graphs to describe motion  …our skills in interpreting graphs of motion  …our skills in describing motion using physics terms correctly  … our understanding of using numbers and equations to describe motion

Time (s) speed (m/s) Example 1 (a)Describe the motion shown on the speed time graph. (b)Calculate the acceleration for each part of the graph. (c)Find the distance travelled in the first 4 seconds.

Time (s) Velocity m/s Example 2 Draw an acceleration time graph Find the maximum displacement from the start. Find the final displacement.

Sketch graphs a v t t See summary notes page10

t/s Draw a velocity time graph for the first 8s of motion. a / ms t/s v/ms Now do tutorial questions - Displacement time graphs – all - Velocity time graphs Qu 1 to 3 - Acceleration time graphs Qu 1,2

© Nicola Jones

© Erich Schrempp / Science Photo Library For no air resistance

Describe the motion in detail using the words velocity, acceleration and displacement.

Sketch graphs to show how the speed, distance, velocity and displacement vary with time during the free-fall. Time Speed Time Distance Time Velocity Time Displacement Remember up is positive

Time (s) Speed (m s –1 ) Displacement (m) Velocity (m s –1 ) Distance ( m) Time (s) Remember up is positive

Sketch the acceleration time graph for a falling ball. 0 a ( m s –2 ) Time (s)

Consider the ball being dropped, allowed to bounce and return to its original height. Sketch your predictions for speed–time, velocity–time and acceleration–time graphs.

Describe the motion. 0

When dropped, the ball gains speed in the negative direction hence the –ve sign for acceleration. The ball then loses speed in the positive direction, coming to rest at the original height. Does this happen in real life? Explain! 0

0 a ( m s –2 ) Time (s)

In real life energy is changed to heat and sound. ics.co.uk/bouncing- ball.html

Ball falling from rest – up direction is positive v t What would the acceleration time graph look like? What would the displacement time graph look like? Notes: All red lines have same gradient – (on Earth this will be – 9.8 m/s 2 as this is acceleration due to gravity). Above the time axis the ball is moving upwards, below it is moving downwards Now look at the graphs in your summary notes page 11.

Consider a tennis ball thrown upwards and allowed to fall back to its starting position.

Sketch the velocity, speed and acceleration graphs to describe its motion until it returns to its starting position. Tutorial questions p11 Qu 4, p 13 Qu 5

Estimate the maximum velocity of a jumping popper.

What assumptions and estimations did you make? How could your calculation be improved? How could you use this to calculate the initial acceleration? Show all your calculations to find the maximum velocity of a jumping popper.

……our understanding of using graphs to describe motion ……our skills in interpreting graphs of motion ……our skills in describing motion using physics terms correctly.