Position: when an object moves, it’s position changes. The reference point in which the object moved from is referred to as the origin point. For example, say that you moved 3km east of Deakin University. The position is, for example, Hungry Jacks, but the origin is Deakin University. For motion along a line, we can choose the line of motion to be the x -axis of a coordinate system. The origin is the point x = 0, in our case that is Deakin University. The position of an object can be described by its x -coordinate, which tells us both how far the object is from the origin and on which side. When we moved to Hungry Jacks, we moved + x -direction to the east. Then x = + 3 km. However, if we went 30km west, x= -30.

West Doncaster Origin (Deakin University) Hungry Jacks East

X= -30 X=0 X=3

Distance: is a length of measurement between two points. For example, to get from Deakin University to Hungry Jacks is 3km. It does not have a direction (e.g. east) and thus it is not a vector rather a scalar (a single number with a unit).

Displacement: is the change of the position of an abject; the final position minus the initial position. The displacement is written Δ x, where the symbol means the change in the quantity that follows. X f is the final position and x i is the initial position (origin).

Δ x (In a straight line) = x f − x i

Displacement has two aspects: magnitude and direction. The magnitude is measured in the same units as distance (metres) & direction is the refers to which way it goes such as up, down, left, east etc. A quantity that has both a magnitude and direction is referred to

as a vector.

What if change in position did not occur in a straight line?

Worked Example: A bird flies 3km east then 3 km north.

3km

(3)

a) What is the distance it travelled?

3km + 3km = 6km

b) What is its displacement at the end of the journey?

An object's displacement vector is drawn as an arrow from the object's initial position to its final position, regardless of the actual path followed between these two points. The length of the vector arrow is proportional to the magnitude of the vector quantity.

You can then use Pythagoras' theorem (C² = A² + B²) to find the length of the displacement vector.

C2 = 3² + 3² = 9 + 9 = 18 C = √18 = 4.2 km

Therefore, the displacement at the end of the journey is 4.2 km NE (don't forget a vector needs a direction)

Worked example 2: a marathon runner completes a lap.

8km

a) What is the distance?

8km

b) What the displacement at the end of the lap?

Since the marathon runner returns to his original place and thus isn’t ‘displaced’, the displacement= 0km.

Speed: is the distance travelled per unit of time and is always a positive scalar. The average speed is defined as the distance travelled divided by the time elapsed (Δt).

Speed= !"#$%&'(

∆$ (',%&-( "& $".()

(4)

For example, if a car travelled 90km in one hour, then the speed is 90km/h.

Velocity: the speed of an object in a given direction. It is a quantity with magnitude and direction. The magnitude is the speed with which the object moves and the direction is the direction of motion.

Velocity (Vx)= 0 1 (',%&-( "& 2"#$%&'()

0 $ (',%&-( "& $".() or ₍₁⁽³⁴⁵³⁶⁾

451₆)

Note: displacement indicates by how much and in what direction the position has changed, but implies nothing about how long it took to move from one point to the

other. Velocity depends on both the displacement and the time interval.

Worked example 3: who is running faster, A or B? (dot represents position of person) A O O O O O O O O O O O

B O O O O O

As the same time has elapsed between each particle and participant, the decreasing distance between the particles shows that the object is slowing down (i.e. less distance travelled in the same amount of time). Therefore, we can say that B is faster.

The position-versus-time graph: used to represent linear motion using a graph.

x= time and y= distance. The motion graph shows that the object moves from a starting position (x = 0) at a time we choose to call t = 0 min. It moves 200 m in 3 minutes. Then there is a three-minute period in which the distance travelled during each time interval becomes shorter.

Then at t = 6 min, the distances travelled within each interval are longer. The speed changes and by finding the slope (⁽³₍₁⁴⁵³⁶⁾

451₆)), we can calculate the velocity. For example, between 6min and 9min (3mins total) the object moves a total of 180m and therefore, the object moves at a speed of 60m/m.

0 200 400 600 800

0 2 4 6 8 10

(5)

Acceleration: describes how much an object's velocity v changes in a given amount of time (t). Acceleration is a vector and requires a magnitude and a direction. For example, the rollercoaster in Gold Coast goes from 0km/h to 110km/h in three seconds. First, we must convert this to meters per second, which is 30.6 m/s. Then, using the equation below, we can work out the acceleration. 30.6/4= 7.6. The acceleration therefore is 7.6m/s2.

Acceleration = 07 (',%&-( "& 7(89'"$3 )

0 $ (',%&-( "& $".() or ^71:571;_$:5$;

Kinematics: calculating motion without referencing force. The following below are equations if acceleration is constant and motion occurs along a straight line.

• vf= vi + aDt

• xf= xi + viDt + ^;_:a(Dt)²

• vf2= vi2 + 2aDx

• Dx= ^;_: (vf + vi) Dt

Free fall: same equations as above. G (acceleration)= 9.80m/s². It is negative if an object is moving down and positive if an object is being thrown up. You can use the above kinematic equations, just substitute g for a.

Week 1 Questions

1. A rock is thrown straight up with a speed of 25.0 m/s. What is the maximum height reached by the rock?

Dt = _%^<

= >.@A (2B( $9 -C%<"$3) ^:=

= 2.55s

25(velocity)= m/s 25=m/2.55

m= 63.78m (this is the distance of x, the whole round trip) h= 63.78/2(half-way up)

h= 31.9m

2. A Pokemon ball is thrown straight up with an initial speed of 18.9 m/s. What is its speed after 7.5 s?

Table of Contents

Table of Contents

Week One: ‘Motion Along a Line’