Unit 1: Forces and Motion: Chapter 1: Movement and Position

Movement and Position

It is very useful to be able to make predictions about the way moving objects behave.

Learning Objectives

• Plot and explain distance-time graphs.

• Know and use the relationship between average speed, distance moved, and time taken: average speed = distance moved / time taken

• Practical: investigate the motion of everyday objects such as toy cars or tennis balls

• Know and use the relationship between acceleration, change in velocity and time taken: acceleration = change in velocity / time taken $a = (v - u) / t$

• Plot and explain velocity-time graphs.

• Determine acceleration from the gradient of a velocity-time graph.

• Determine the distance traveled from the area between a velocity-time graph and the time axis.

• Use the relationship between final speed, initial speed, acceleration, and distance moved: $(final speed)^2 = (initial speed)^2 + (2 \times acceleration \times distance moved)$
  $v^2 = u^2 + (2 \times a \times s)$

Units

• Kilogram (kg) as the unit of mass, meter (m) as the unit of length, and second (s) as the unit of time. Convert units in calculations to the base units of kg, m, and s when you meet these subdivisions and multiples.

• Speed and velocity: meter per second (m/s)

• Acceleration: meter per second squared (m/s²).

• Force: newton (N)

• Gravitational field strength: newton per kilogram (N/kg)

Average Speed

• Average speed, $v = distance moved, s / time taken, t$

Units of Speed

• Meters per second (m/s).

• Kilometers per hour (km/h).

• Centimeters per second (cm/s).

• Miles per hour (mph).

• Kilometers per hour (kph or km/h).

Rearranging the Speed Equation

• $distance moved, s = average speed, v \times time, t$

• $time taken, t = distance moved, s / average speed, v$

Speed Trap Experiment

1. Measure 50 m from a start point along the side of the road.

2. Start a stopwatch when your partner signals that the car is passing the start point.

3. Stop the stopwatch when the car passes you at the finish point.

• The average speed of the car during the journey is the total distance traveled divided by the time taken for the journey.

• The speedometer shows the instantaneous speed of the car.

• $average speed, v = distance moved, s / time taken, t$

• If the time measured is 3.9 s, the speed of the car in this experiment is: average speed, $v = 50m / 3.9s = 12.8 m/s$

Distance-Time Graphs

• Plot a graph of the distance traveled (vertical axis) against time (horizontal axis).

• The distance-time graph tells us about how the car is traveling.

• Constant speed is shown immediately by the fact that the graph is a straight line.

• The slope or gradient of the line tells us the speed of the car - the steeper the line the greater the speed of the car.

• $speed = gradient = distance / time = 30 m / 2.5s = 12 m/s$

Displacement

• Displacement is distance traveled in a particular direction.

• Vector quantities have magnitude (size) and a specific direction.

• $average velocity = increase in displacement / time taken$

Example 1

• The global positioning system (GPS) shows two points on a journey. The second point is 3 km north-west of the first.

• A walker takes 45 minutes to travel from the first point to the second. Calculate the average velocity of the walker.

  • increase in displacement is 3 km north-west

  • time taken is 45 min (45 min = 0.75 h)

  • average velocity = increase in displacement / time taken = $3 km / 0.75 h = 4 km/h$ north-west

• The walker has to follow the roads, so the distance walked is greater than the straight-line distance between A and B (the displacement). The walker's average speed (calculated using distance) must be greater than his average velocity (calculated using displacement).

Activity 1

• Investigate the motion of everyday objects such as toy cars or tennis balls

• Use simple apparatus to investigate the motion of a toy car.

• Measure the average speed, v of the car for different values of h.

Practical Investigation

• Measure the height, h, of the raised end of the wooden track.

• To find the average speed use the equation: average speed, $v = distance moved, s / time taken, t$

• Measure the distance AB with a meter rule and measure the time it takes for the car to travel this distance with a stop clock.

• To make these smaller the time to travel distance AB should, for a given value of h, be measured at least three times and an average value found.

• Always start the car from the same point, A.

• If one value is quite different from the others it should be treated as anomalous (the result is not accurate) and ignored or repeated.

• The conclusion you draw must be explained with reference to the graph, for example, if the best fit line through the plotted points is a straight line and it passes through the origin (the 0, 0 point) you can conclude that there is a proportional relationship between the quantities you have plotted on the graph.

Photographic Methods

• Carry out the experiment in a darkened room using a stroboscope to light up the object at regular known intervals (found from the frequency setting on the stroboscope) with the camera adjusted so that the shutter is open for the duration of the movement.

• Using a video camera and noting how far the object has traveled between each frame - the frame rate will allow you to calculate the time between each image.

Acceleration

• Acceleration is the rate at which objects change their velocity.

• $acceleration, a = final velocity, v - initial velocity, u / time taken, t$

• $a = (v - u) / t$

• The direction in which the acceleration occurs is important as well as the size of the acceleration.

• Velocity is measured in m/s, so increase in velocity is also measured in m/s.

• Acceleration, the rate of increase in velocity with time, is therefore measured in m/s/s (read as 'meters per second per second"). We normally write this as m/s² (read as 'meters per second squared').

Example 2

• A car is traveling at 20 m/s. It accelerates steadily for 5 s, after which time it is traveling at 30 m/s. Calculate its acceleration.

  • initial or starting velocity, $u = 20 m/s$

  • final velocity, $v = 30 m/s$

  • time taken, $t = 5s$

  • $a = (v - u) / t = (30 m/s - 20 m/s) / 5s = 10 m/s / 5s = 2 m/s^2$

• The car is accelerating at 2 m/s².

Deceleration

• Deceleration means slowing down.

• A decelerating object will have a smaller final velocity than its starting velocity.

• A negative acceleration simply means deceleration.

Example 3

• An object hits the ground traveling at 40 m/s. It is brought to rest in 0.02 s. What is its acceleration?

  • initial velocity, $u = 40 m/s$

  • final velocity, $v = 0 m/s$

  • time taken, $t = 0.02 s$

  • $a = (v - u) / t = (0 m/s - 40 m/s) / 0.02 s = -40 m/s / 0.02 s = -2000 m/s^2$

• In Example 3, we would say that the object is decelerating at 2000 m/s².

Measuring Acceleration

• Galileo was interested in how and why objects, like the ball rolling down a slope, speed up, and he created an interesting experiment to learn more about acceleration.

• Galileo wanted to discover how the distance traveled by a ball depends on the time it has been rolling.

• By adjusting the positions of the bells carefully t is possible to make the bells ring at equal intervals of time as the ball passes.

• Galileo noticed that the distances traveled in equal time intervals increased, showing that the ball was traveling faster as time passed.

Velocity-Time Graphs

• A graph showing how the velocity of the ball is changing with time is called a velocity-time graph.

• The graph is a straight line, then the acceleration is uniform.

More about Velocity-Time Graphs

• The slope or gradient of a velocity-time graph is found by dividing the increase in the velocity by the time taken for the increase.

• Increase in velocity divided by time is the definition of acceleration.

• Measure the acceleration of an object by finding the slope of its velocity-time graph.

• The steeper the slope of the air-track the greater the glider's acceleration.

Area Under a Velocity-Time Graph Gives Distance Traveled

• The area under a velocity-time graph is equal to the distance traveled by (displacement of) the object in a particular time interval.

• Average velocity = initial velocity + final velocity / 2 = (u+v) / 2

• Find the distance traveled for more complicated velocity-time graphs by dividing the area beneath the graph line into rectangles and triangles.

Equations of Uniformly Accelerated Motion

• $a = (v - u) / t$

• $v - u = a \times t$

• $v = u + at$

Example 4

• A stone accelerates from rest uniformly at 10 m/s² when it is dropped down a deep well. It hits the water at the bottom of the well after 5 s. Calculate how fast it is traveling when it hits the water.

  • initial velocity, $u = 0 m/s$

  • acceleration, $a = 10 m/s^2$

  • time, t, of the acceleration = 5s

  • $v = 0 m/s + (10 m/s^2 \times 5 s) = 50 m/s$

• The stone hit the water traveling at 50 m/s (downwards).

Uniformly Accelerated Motion

• The equations of uniformly accelerated motion will give you correct answers when solving any problems that have objects moving with constant acceleration.

• Examples in which objects accelerate or decelerate (slow down) at a constant rate often have a constant acceleration due to the Earth's gravity (which we take as about 10 m/s²).

• Objects only fall with constant acceleration if we ignore air resistance and the distance that they fall is quite small.

• These equations of uniformly accelerated motion are often called the 'suvat' equations.

• $(final speed)^2, v^2 = (initial speed)^2, u^2 + (2 \times acceleration, a \times distance moved, s)$

• $v^2 = u^2 + 2as$

Example 5

• A cylinder containing a vaccine is dropped from a helicopter hovering at a height of 200 m above the ground. The acceleration due to gravity is 10 m/s². Calculate the speed at which the cylinder will hit the ground.

  • acceleration, $a = 10 m/s^2$

  • distance, $s = 200 m$

  • initial velocity, $u = 0 m/s$

  • $v^2 = u^2 + 2as = 0 m/s^2 + (2 \times 10 m/s^2 \times 200 m) = 4000 m^2/s^2$

  • $v = \sqrt{4000 m^2/s^2} = 63.25 m/s$

Syllabus

• Plot and explain distance-time graphs

• Know and use the relationship between average speed, distance moved and time taken:

  • average speed = distance moved / time taken

• Practical: investigate the motion of everyday objects such as toy cars or tennis balls

• Know and use the relationship between acceleration, change in velocity and time taken:

  • acceleration = change in velocity / time taken

  • $a = (v - u) / t$

• Plot and explain velocity-time graphs

• Determine acceleration from the gradient of a velocity-time graph

• Determine the distance traveled from the area between a velocity-time graph and the time axis

• Use the relationship between final speed, initial speed, acceleration and distance moved:

• $(final speed)^2 = (initial speed)^2 + (2 \times acceleration \times distance moved)$

• $v^2 = u^2 + (2 \times a \times s)$

Formulae:

average speed = distance moved / time taken
$a = (v - u) / t$
$(final speed)^2 = (initial speed)^2 + (2 \times acceleration \times distance moved)$
$v^2 = u^2 + (2 \times a \times s)$
Average speed, $v = distance moved, s / time taken, t$
$distance moved, s = average speed, v \times time, t$
$time taken, t = distance moved, s / average speed, v$
$speed = gradient = distance / time$
$average velocity = increase in displacement / time taken$
Average speed, $v = distance moved, s / time taken, t$
$acceleration, a = final velocity, v - initial velocity, u / time taken, t$
$a = (v - u) / t$
$v = u + at$
$(final speed)^2, v^2 = (initial speed)^2, u^2 + (2 \times acceleration, a \times distance moved, s)$
$v^2 = u^2 + 2as$
$a = (v - u) / t$
$(final speed)^2 = (initial speed)^2 + (2 \times acceleration \times distance moved)$
$v^2 = u^2 + (2 \times a \times s)$