CONTENTS  Scalars and Vectors.  Bearings & Compass headings.  Drawing Vectors.  Vector addition & subtraction.  Relative velocity.  Change in velocity.  Exercises from Rutter.  Homework.

Scalars and Vectors Every quantity can have two types of value. They can either be: a.SCALAR or b.VECTOR But what does this mean? SCALAR quantity just gives the MAGNITUDE VECTOR quantity gives MAGNITUDE & DIRECTION

DistanceDisplacement SpeedVelocity Mass Energy Density Acceleration Force Momentum

N E 4 Km East + 3 Km North R = ?  = Bearing of Answer = 5 km o or 5 km E53.1 o N

BEARINGS: These can be written in two forms as: a.Bearing which is a proportion of a circle -Three sig.figs. E.g. 53 o written as 053 o. Move from north 53 o round clockwise. b. As a compass heading -Start with direction pointing -Place number of degrees from 0 – 90 o -Place compass direction you are moving in after the number of degrees. E.g. 53 o west of north can be rewritten N53 o W

What is the resultant force R ? 6N 8N We ADD vectors HEAD to TAIL R  6N Answer = 10N N53.1 o E or 10N o

à 1. First draw vector Ã Ñ 2. Then draw vector Ñ so that the tail of vector Ñ touches the head of vector à 3. The resultant is the vector that starts at the tail of vector à and ends at the head of vector Ñ. Ã Ñ Ã + Ñ Two vectors à & Ñ are added together = à + Ñ Ã +Ñ

Example 1: A tourist couple visited several places of interest around Dunedin. They began by travelling 5.0km east from their hotel, then 3.0km south, then 4.0km west and finally 6.0km north. What was their final displacement, Ỹ, from the hotel? SOLUTION: Displacement is a vector quantity so the four displacements covered by the tourists can be represented by vectors and added using a vector diagram. The resultant is their final displacement from the hotel. 5.0km 3.0km 4.0km 6.0km Ỹ  N SCALE

In the north/south direction the couple have moved 3.0km S & 6.0km N = 3.0km N In the east/west direction they have moved 5.0km E & 4.0km W = 1.0km E Resultant found by adding two vectors together. Ỹ =  ( ) =  10 = 3.2km tan  = 3.0/1.0 = 3.0  = 72 o Ans = 3.2km, 72 o N of E OR 018 o N E  3.0km N Ỹ 1.0km E

Vector Subtraction If any negative sign is used then this means that the direction has changed to the opposite direction. Two vectors à & Ñ are subtracted = à - Ñ Ã -Ñ Ã 1. Begin with à -Ñ 2. Add -Ñ to give3. ….the resultant, à - Ñ. à -Ñ Ã - Ñ

To find the change in velocity,  v, from an initial velocity v i to a final velocity v f, the vector subtraction v f – v i is done. vfvf vivi vfvf -v i vv  v = final velocity – initial velocity  v = v f – v i  v = v f + (-v i )

RESOLUTION OF VECTORS - COMPONENTS F What is the component of the force F in the x - direction? x Dropping a perpendicular  C C is the component of F in the x - direction Hence C = F cos 

F What is the component of the force F in the y - direction? y Dropping a perpendicular  E is the component of F in the y - direction Hence E = F sin  E x

Example 2: The vertical/horizontal reference frame (a) shows a vector representing a 10.0N force. The direction of the vector is angled up from the horizontal at 30 o. The two force vectors which represent the components of this 10N force are found by drawing two construction lines, one parallel to the vertical axis and the other parallel to the horizontal axis. The components are labelled (b). The 10N vector and its components forma right angle triangle (c). Use trigonometry to calculate the vertical & horizontal components. 30 o 10N vertical horizontal 30 o vertical horizontal 10N 30 o y x a. b.c.

SOLUTION: Vertical component Sin 30 o = opposite/hypotenuse Sin 30 o = y/10 y = 10sin30 o y = 5.00N Horizontal component Cos 30 o = adjacent/hypotenuse Cos 30 o = x/10 x = 10cos30 o x = 8.66N This means that the 10N force at 30 o is made up of a horizontal component of 8.66N and a vertical component of 5.00N From this the 10N force at 30 o has a horizontal effect of 8.66N and a vertical effect of 5.00N

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RELATIVE VELOCITY This is the velocity of one object in relation to another object. The velocity of an object may appear to be different on where it is measured from. What do you notice about the motion of each of the planes relative to a person watching from the ground?

Example 3: A motorbike moves forward at 100kmh -1 and beside it is a jeep moving at 50kmh -1 in the same direction. How does fast does the bike appear to be moving observed by the jeep driver? SOLUTION: Bike moves forward fro the jeep at = 50kmh -1. To the driver of the jeep it is though the jeep is stationary and the bike is moving at 50kmh kmh -1  100kmh -1 

Example 4: A tank rides horizontally at 50kmh -1 through the rain. The steady rain falls vertically at 15kmh -1. What is the velocity of the rain relative to the tank? Rain 15kmh -1 Tank 50kmh -1 SOLUTION: The two velocities are both given relative to the ground. Yet experience tells us that relative to the tank the rain would appear to be coming towards their face than straight down. The velocity of the rain relative to the tank is found by subtracting the tanks velocity from the velocity of the rain.

Using Pythagorus (where v is the magnitude): v =  = 52kmh -1 Using trigonometry  = tan -1 (15/50)  = 17 o 52kmh -1 at an angle of 17 o to the horizontal  50kmh -1 -v tank 15kmh -1 -v rain V rain rel tank

Observe the motion of the river boats below. In each case, the boat is heading across the river. In the top case, the presence of a current causes the boat to travel downstream as it simultaneously moves across the river; the result is that it reaches the opposite shore at a point downstream from where it started. In the bottom case, the absence of a current means that the boat moves straight across the river; it ultimately reaches a point on the opposite shore directly across from where it started. It is important to note that the current is what carries the boat down the stream; it is not the motor of the boat that provides its downstream motion.

Example 5: A boat starts to cross a river which is 2.0km wide. The boat travels at 20kmh -1 relative to the surface of the water and points directly across the river. The river is moving 6.0kmh -1 downstream relative to the river bank. a. What is the boat’s velocity relative to the river bank? b. How long does the boat take to cross the river? c. At what point does the boat land on the other side? 20kmh -1 relative to the water’s surface 20kmh kmh -1 relative to the bank

SOLUTION: The current is relative to the river bank and the river bank (and any observer standing on the bank) is at rest. a.The boats velocity relative to the river bank is the combined result of its velocity relative to the river and the velocity of the river relative to the river bank: v boat rel river + v river rel bank = v boat rel bank   20kmh kmh -1 v boat rel bank Using Pythagorus and trigonometry, the boat moves with a speed of: v =  ( ) v = 21kmh -1 tan  = 20/6.0  = 73 o V boat rel bank = 21kmh -1 at 73 o

b. The boat continues to travel across the river at 20kmh -1, despite the fact that it is also being carried downstream. The river’s motion does not affect the boat’s movement across the river because the motion of the river is at right angles to the direction in which the boat is pointed.  time taken to cross the river is t = d/v = 2.0/20 = 0.10hours (6minutes) c. As well as travelling across the river, the boat is acrried at 6.0kmh -1 downstream for 0.10 hours (the time taken to cross the river). The distance travelled downstream is:  distance d = vt = 6.0 x 0.10 = 0.60km The boat touches the opposite river bank 0.60km downstream from a point directly across from the river.

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