Distance, Speed and Acceleration Graphs

DISTANCE, SPEED AND ACCELERATION GRAPHS AMD THEIR VECTOR EQUIVALENTS

We will consider motion of a particle in one dimension i.e. back and forth along a straight line.

Activity 1 Class discussion about the differences between speed and velocity, distance and displacement. Demonstrate these differences by using the room and pacing about.

Distance and Speed are scalar quantities because they just have a size or magnitude Displacement and Velocity are their vector equivalents that have both size and direction. Since we are working in 1-d and can only go up /down or left/right the direction will be indicated by a negative sign on the velocity or displacement. Acceleration can either be a scalar or a vector there isn’t a different word for the vector quantity of acceleration

1. Distance, Displacement and Position Graphs Distance is how far something has travelled. It always increases (like the odometer in a car) Displacement is the distance, measured in a straight line, of an object from its initial position. The displacement is not necessarily the same as the distance travelled.

Activity 2 Consider the distance/time graph (Slide 2 – Lesson 1 and 2) Give the class 10 minutes to answer the questions on Slide 3 Go thorough the answers getting the students to explain their answers. Then show Side 3 which overlays the displacement time graph for the journey over The distance time graph. Discuss which of the answers to the questions are more easily obtainable and which are less easily obtainable

The gradient of a distance/time graph tells you the speed the object is travelling at. The gradient of a displacement/time graph tells you the velocity that the object is travelling at. The faster the subject is moving the steeper the gradient. (Note we can estimate the velocity or speed on a graph that curves by using a tangent to find the gradient at a point)

Example 1 : A girl throws a ball from head height (1.5m above the ground), vertically into the air. The ball rises to a maximum height of 10m, before landing on the ground at her feet. Find the distance travelled by the ball and its displacement. distance travelled= 8.5 + 10 = 18.5m displacement= 1.5m If the girl had instead rolled the ball along the ground, the distance travelled and the displacement would have been equal.

Example 2 : A man walks four miles due East, followed by three miles due North. Find the distance travelled and the displacement. distance travelled= 4 + 3 = 7 miles

2 2 5 displacement= 4 + 3 = 5 miles 3

In other words, displacement is only concerned with where an object is, whereas distance is concerned with how far it travelled getting there. Displacement is a vector quantity - it has both magnitude and direction. For one-dimensional motion, this means the displacement can be positive or negative. In the diagram below, object A has a displacement of −3 units (measured from the origin O), and object B has a displacement of 2 units. Note that distance is a scalar quantity, and therefore the distance of A from O is simply 3 units.

x A O B

Activity 3 Trolley Graph Match SWF activity

Sketch displacement/time and distance/time graphs for the following situations:

1) The pendulum of a grand father clock that oscillates 5 cm to either side and takes 1 second to swing from one side to the other 2) The height of a cricket ball thrown vertically upwards from a height of 1m to a height of 6m if it lands on the floor after 4 seconds 3) The height of a lift as it leaves the ground floor, takes 30 seconds to reach floor 5, waits for 1 minute and then takes 40 seconds to descend to the basement 4) The height of a bungee jumper as they fall from a bridge 50m high.

Activity 4 Get the students to describe their own situations for distance/time and displacement time graphs. Sketch each others showing the differences in the two types of graph. Activity 5 Show slide 10 and discuss the differences between the three graphs.

A Position/time graph is the same as a displacement/time graph except that the displacements are not necessarily measured from the starting point of the object. Instead an origin of your choosing is used and the objects displacement from that point is shown on the graph. (Note many people still consider this a displacement/time graph as it shows much the same information)

Exercise 1B questions 2 to 4 pg 9

2. Speed and Velocity graphs Velocity is the rate of change of displacement of an object, whereas speed is the rate of change of distance. Both are measured in ms-1 (metres per second). A velocity of 5ms-1 means that an object will travel 5 metres in every second. Velocity is a vector quantity (and therefore in one-dimensional motion can be positive or negative), whereas speed is a scalar quantity. In the diagram below, A and B both have speed 2ms-1. The velocity of A is −2ms-1, since it is travelling in the negative direction, and the velocity of B is 2ms-1, since it is travelling in the positive direction. 2ms -1 2ms -1

x A O B

The gradient of a speed/time graph tells you the acceleration (as a scalar) of the object at that instant. The gradient of a velocity/time graph tells you the acceleration (as a vector) of the object at that instant The faster the subject is moving the steeper the gradient. (Note we can estimate the acceleration on a graph that curves by using a tangent to find the gradient at a point) The area under a speed/time graph between two times tells you the distance travelled in that time. The area under a velocity/time graph between two times tells you the displacement of the object in that interval (using the start time as the origin)

Activity 6 Show slides 3 to 9 and practice interpreting and drawing speed/time graphs. The last slide branches into velocity/time graphs as an extension (plenary) Note that to change a velocity/time graph into a speed/time graph you just need to reflect any negative velocities into the positive part of the graph. Note also that negative velocities mean that the object is travelling in the opposite direction.

If the velocity is changing smoothly from negative to positive or vice-versa (ie crossing the time axes) it means that the object is momentarily (we tend to say instantaneously) at rest and is changing its direction of travel. At this point the displacement will have reached a localised maximum or minimum value.

Activity 7 Interpreting Velocity/time graphs. Give pairs a copy of all the graphs in this power point and a list of the questions

Example 1 : A cyclist leaves home O and rides along a straight road with a constant acceleration. After 10 seconds, he has reached a point A with a speed of 15ms-1, and he maintains 51234 05 51234 05 this speed for a further 20 seconds, until he reaches B, before retarding uniformly to rest at C. The whole journey takes 45 seconds. Sketch a speed-time graph for the journey and find a) his acceleration for the journey from O to A. b) his retardation for the journey from B to C. c) the total distance travelled from O to C.

v A B 15

O 5 10 15 20 25 30 35 40 45 t a) The acceleration is the gradient of the graph from O to A. 15 acceleration = 10 =1.5 ms-2 b) The acceleration is the gradient of the graph from B to C. -15 acceleration = 15 = -1 ms-2 So the retardation is 1ms-2. c) The total distance travelled is the area under the graph.

1 distance=2 (20 + 45) 15 = 487.5 m Example 2 : A car travelling along a straight road passes point A when t = 0, and maintains a constant speed until t = 24 seconds. The driver then applies the brakes and the car retards uniformly to rest at point B. Before the brakes were applied the car had 4 travelled 5 of the total distance AB. Sketch the speed-time graph and calculate the time taken for the car to travel from A to B. v A P u

B O 24 T t Let the initial speed be u, and the time taken for the journey be T. distance travelled at constant speed= 24u 1 total distance travelled=2 (24 +T ) u So we have

4 1 24u= 5� 2 (24 T ) u 2 24=5 (24 +T ) 60= 24 +T T = 36 seconds The time taken is 36 seconds. There is a much quicker, geometrical solution to this problem. v x 4x x O 24 T t The dotted rectangle has half the area of the rectangle to the left, and therefore half the width. Its width must therefore be 12, and so T is 36.

Example 3 : A man is jogging along a straight road at a constant speed of 4ms-1. He passes a friend with a bicycle who is standing at the side of the road and 20 seconds later the friend starts cycling to catch him up. The cyclist accelerates at a uniform rate of 3ms-2 until he reaches a speed of 12ms-1. He then maintains a constant speed. a) On the same diagram, sketch the speed-time graphs for the jogger and cyclist. b) Find the time that elapses before the cyclist reaches the jogger. a) Let the cyclist reach the jogger at time T.

12 cyclist

4 jogger

20 24 T t b) We require the distance travelled by the jogger to be equal to the distance travelled by the cyclist at time T. So the areas under the two graphs up to time T must be equal. 1 4T=2 �[ ( T + 24) - ( T 20)] 12 4T= 6(2 T - 44) 4T= 12 T - 264 264= 8T T = 33 seconds

Example 4 : Two stations A and B are a distance of 6x metres apart along a straight track. A train starts from rest at A and accelerates uniformly to a speed v ms-1, covering a distance x metres. The train then maintains this speed until it has travelled a further 3x metres. It then retards uniformly to rest at B. Make a sketch of the velocity-time graph for the motion and show that if T is the time taken for the train to travel from 9x A to B, then T = seconds. v v 3x

3x x 2x B A T t 3x The time taken for the steady speed phase is clearly . v Using the formula for the area of a trapezium,

1 area=2 (a + b ) h

1 骣3x 6x=琪 + T v 2 桫v 12x= 3 x + Tv 9x= Tv 9x T = v

Alternatively, we can solve the problem v geometrically. Comparing expressions for the area of the v rectangle, x 2x Tv= 9 x 3x 9x x 2x B T = A v T t

Example 5 : Now solve example15 4 from section 8 the easy way!

T t 1 area of triangle = 2 bh 1 60= 2 创T 15 120= 15T T = 8 seconds Notice how much easier this is, and that the retardation in the second phase is not even needed!

Exercise 1D questions 1 to 8 pg 16

3. Acceleration/time graphs Acceleration is the rate of change of velocity of an object. It is measured in ms-2 (metres per second squared). An acceleration of 5ms-2 means that the velocity of an object will increase by 5ms-1 in every second. Similarly, an acceleration of −5ms-2 means that the velocity of an object will decrease by 5ms-1 in every second. An acceleration of −5ms-2 is often known as a retardation of 5ms-2.

Example 1 : An object is travelling with a velocity of 3ms-1. It is accelerating at 2.5ms-2. Find its velocity after 7 seconds. velocity= 3 + (7 2.5) = 20.5 ms-1

Example 2 : An object is travelling with a velocity of 2ms-1. It is accelerating at −1.5ms-2. Find its velocity after 4 seconds. velocity= 2 + (4� 1.5) = -4 ms-1

Activity 8 Instead of using the examples below use the excel spreadsheet displacement+velocity+acceleration locator to test pupils understanding of the vector quantities for acceleration, velocity and displacement.

Example 3 : Draw a diagram to illustrate an object for which... s : negative v : positive a : positive

O x speeding up Example 4 : Draw a diagram to illustrate an object for which... s : positive v : positive a : negative

O x slowing down

Example 5 : Draw a diagram to illustrate an object for which... s : positive v : negative a : negative

O x speeding up It may seem strange that an object with negative acceleration is actually speeding up. This happens because the object is moving with negative velocity, and decreasing a negative velocity actually increases its magnitude (i.e. its speed). For example, if the velocity is −3ms-1 and the acceleration is −1ms-2, then one second later the velocity will be −4ms-2, which is faster than −3ms-2.

Example 6 : Draw a diagram to illustrate an object for which... s : negative v : positive a : zero

O x constant speed

Example 7 : Draw a diagram to illustrate an object for which... s : negative v : zero a : positive

O x not moving, about to move to the right

Example 8 : Draw a diagram to illustrate an object for which... s : zero v : negative a : positive

The area under an acceleration/timeslowing graph down betweenO two times tells you thex increase/decrease in velocity between those times. Note that when acceleration is zero velocity is at a localised maximum or minimum value.

Exercise 1C questions 1 to 6 pg 11 12340 1234

4. Uniform Velocity A particle travels with uniform velocity if its speed and direction are constant. The graph of displacement against time is a straight line, and the gradient is the velocity.

1 2 3 4 t 10 velocity= = 5 ms-1 2

5. Average Velocity The velocity of an object may change over time. We define displacement average velocity = time taken distance average speed = time taken

Example 1 : A car travels for 4 hours at 100km/h in one direction, and then for 160km at 80km/h in the opposite direction. Find the average velocity and the average speed of the car. In the first stage of the motion, the car travels 400km. 160 The second stage of the motion takes = 2 hours. 80 displacement average velocity = time taken 400- 160 = 6 = 40 km/h 1234560 0 distance average speed123456 = time taken 400+ 160 = 6 1 = 933 km/h

We can interpret these figures thus. If the car is to get to its s final destination by travelling in a straight line, it should move 400 at a constant 40 km/h. If the car is to get to its final destination by its actual route, it should move at a constant 300 1 93 3 km/h. 200 In the displacement-time graph opposite, the average velocity 100 is the gradient of the dotted line. It is not the average of the 12340 individual velocities. 1234 1 2 3 4 5 6 t

6. Uniform Acceleration An object travels with uniform acceleration if its acceleration is constant. The graph of velocity against time is a straight line and the gradient is the acceleration.

1 2 3 4 t 10 acceleration= = 5 ms-2 2