Take A Break! What about this??? Which one is false?

Take A Break! Aim & Throw where????

Assessment Objectives:
(a) distinguish vector and scalar quantities and give examples of each. (b) add and subtract coplanar vectors. (c) represent a vector as two perpendicular components.

Vectors Scalar quantities are physical quantities that can be represented by a magnitude (size) only. Vector quantities are physical quantities that can only be fully represented by a magnitude and a direction.

Vectors Vectors representation in mechanics: F F ~ AB

Vector Addition For 2 Vectors: Parallelogram Law: For 2 or more vectors: Polygon Method:

Vector Addition a b r

Vector Addition c a b r

Test yourself Now! b a c b c a r c a b r

Vector in equilibrium (to be used in Forces Topic later on)
all the vectors point in the same direction, either clockwise or anti-clockwise all the vectors form a closed polygon a b c d e

Vector Subtraction a – b =a + (-b) a b -b -b a r

Vector Subtraction r = a – b r + b = a a b r

Test yourself Now! eg) The velocity of a car rounding a bend changes from 40 m s-1 due North to 20 m s-1 due East. Find the change in speed of the car. (A) -20 m s-1 (B) -40 m s-1 (C) -60 m s-1 (D) -80 m s-1 Change in speed = = -20 m s-1 (A)

Test yourself Now! eg) Continues from previous eg. Find the magnitude of the change in velocity of the car. (A) m s-1 (B) m s-1 (C) m s-1 (D) m s-1 v=20 m s-1 -u=-40 m s-1 v u=40 m s-1 v v = v – u = v + (-u) v = v + u v=20 m s-1 (C) v = ( ) 1/2 = 44.7 m s-1

Vector Resolution Generally 2 directions perpendicular to each other: Vertically & Horizontally parallel to the plane & perpendicular to the plane

Resolving Vectors F

Resolving Vectors Vectorially: Fy F  Fx

Resolving Vectors Vertically: Fy F Horizontally:  Fx

Test yourself Now! eg) Resolve the following vectors vertically and horizontally: 20 sin 40o 20 N 80 N 40o 20 cos 40o 50 cos 30o 30o 50 cos 30o 50 N

Test yourself Now! eg) What is the component of the block's weight (weight is a vector!) causing it to slide down the slope. 50 cos 30o 30 30 50 sin 30o 50 N (A) 25 N (B) 30 N (C) 35 N (D) 40 N (A)

Kinematics Prepared by : Mr Tan Kia Yen Feb 2009

(a) define displacement, speed, velocity and acceleration. (b) use graphical methods to represent displacement, speed, velocity and acceleration. (c) find the distance traveled by calculating the area under a velocity-time graph. (d) use the slope of a displacement-time graph to find the velocity. (e) use the slope of a velocity-time graph to find the acceleration.

(f) derive, from the definitions of velocity and acceleration, equations which represent uniformly-accelerated motion in a straight line. (g) use equations which represent uniformly-accelerated motion in a straight line, including falling in uniform gravitational field without air resistance. (h) Describe qualitatively the motion of bodies falling in a uniform gravitational field with air resistance. (i) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction.

Introduction This is a branch of mechanics which deals with the description of the motion of objects, without references to the forces which act on the system. In kinematics, we will study the following two types of motion: (a) one dimensional motion, i.e. along a straight line (b) two dimensional motion – projectile motion.

Displacement is the distance travelled along a specific direction.
Definition: Displacement is the distance travelled along a specific direction. vector quantity. symbol is s. SI unit is the metre, m. Displacement is the shortest distance between the initial and final position of the body. Note:

Displacement 10 m Distance = 10 m Displacement ??? 10 m 10 m (i) (f) (i) (f) Displaced 10 m to the left Displaced 10 m to the right

Velocity Definition: Velocity is the rate of change of distance along a specific direction, or simply the rate of change of displacement. vector quantity. SI units are m s-1 symbol is u (initial speed) or v (final speed). <speed> is not always equal to <velocity>. Note:

 Velocity (Quick check!)
A car travels from point A to point B with the following speeds: 20 m s-1 for 2.0 s 40 m s-1 for 2.0 s 60 m s-1 for 6.0 s A B What is the average speed of this car for this journey? A 12 m s-1 C 40 m s-1 B m s-1  D 48 m s-1 <speed> = total distance / time

Then same car now travels back from B to A with the following speeds: A B 60 m s-1 for 4.0 s 40 m s-1 for 6.0 s What is the average speed of this car for the whole journey? A 24 m s-1 C 40 m s-1  B 48 m s-1 D 96 m s-1 <speed> = 480 x 2 / 20

What is the average velocity of this car for the whole journey? A B A 0 m s-1 C 24 m s-1  B 48 m s-1 D 96 m s-1 <v> = displacement / time=0/20

Acceleration is the rate of change of velocity.
Definition: Acceleration is the rate of change of velocity. vector quantity. SI units are m s-2 symbol is a. The acceleration of a body is constant or uniform if its velocity changes at a constant rate. Note: more

Retardation / Deceleration
Retardation or deceleration is a term used to describe a decrease in the magnitude of velocity with time. Occurs when velocity and acceleration are in opposite directions. a v accelerating a v decelerating

Retardation / Deceleration
Are these objects undergoing constant velocity? a change in magnitude e.g. body speeding up or slowing down. (b) a change in both magnitude and direction. (c) a change in direction only e.g. circular motion.

Uniformly accelerated motion
Refers to motion where the velocity of a body is changing at a steady rate. (constant acceleration) Eg) a toy car sliding down a slope a ball being thrown in the air. v t Constant acceleration  no acceleration

Acceleration due to gravity, g ( constant acceleration)
Acceleration due to gravity is produced by the gravitational field of the earth and it is always directed downward relative to the earth’s surface. It can be taken to be constant at 9.81 m s-2 (unless otherwise stated)

Relationship between s, v, a
v = ds / dt (gradient) s =  v dt a = dv / dt (gradient) v =  a dt t s v t v a t s v

Relationship between s, v, a
v = ds / dt (gradient) s =  v dt a = dv / dt (gradient) v =  a dt t s v t v a t v a

instantaneous velocity (gradient of the graph at a particular point)
Displacement-Time Graph Useful information from this graph: (i) instantaneous displacement (the displacement of a body at any instant of time) instantaneous velocity (gradient of the graph at a particular point) average velocity

Displacement-Time Graph
eg) Consider a car traveling along the x-axis. What deduction can be made about the car’s motion if its displacement-time relation is represented by (i) Graph 1 and (ii) Graph 2?

(a) The displacement x carries only positive values. This implies that the car moves along the positive x direction. (b) The displacement x increases uniformly with time, thus the velocity of the car remains constant. (c) Average velocity = <u>=

(a) Since the displacement x remains positive throughout the motion, the car’s positions remain to the right of the origin O. (b) At points A and C, the car is stationary, since its displacement x is not changing at these two points. (c) Around B, the displacement x increases uniformly with time, so the car must be traveling in the positive x-direction with uniform velocity.

(d) After point C, the displacement x of the car is decreasing. This implies that the car is traveling in the opposite direction, i.e. the negative direction. (e) The gradient at D is greater than at E. Thus, the magnitude of the velocity at D is greater than at E. (f) Around E, the displacement x decreases with time at a decreasing rate up to F. This shows that the car eventually stops at F.

Examples of Displacement-Time Graphs
Motion under constant acceleration v = ds / dt s =  v dt a = dv / dt v =  a dt

Motion under constant deceleration

Motion under constant velocity

Velocity-Time Graph Useful information from this graph: (i) instantaneous velocity (the velocity of a body at any instant of time) instantaneous acceleration (gradient of the graph at a particular point) displacement (area under the graph)

Velocity-Time Graph eg) The motion of a body moving along a straight line is given as shown in the graph below. What deductions can be made about the body’s motion between points B and C? v t A B C D E F G H I Area 1 Area 2

(A) Body moving with velocity decreasing at a decreasing rate.
F G H I Area 1 Area 2 Velocity-Time Graph Deductions: (A) Body moving with velocity decreasing at a decreasing rate. (B) Body moving with velocity increasing at a decreasing rate. (C) Body moving with velocity decreasing at a increasing rate. (D) Body moving with velocity increasing at a increasing rate. (B)

Equations of Motion Uniformly accelerated motion refers to motion of a body in which the acceleration is constant. Kinematics equations: v = u + at v2=u2 + 2as Note: They are only valid for cases of uniform acceleration in a straight line. more

Sign Conventions eg) A motorist travelling at 13 m s-1 approaches traffic lights which turn red when he is 25 m away from the stop line. His reaction time (time between seeing the red lights and applying the brakes) is 0.70 s and the condition of the road and his tyre is such that he cannot slow down at a rate of more than 4.5 m s-2. If he brakes fully, what is the total distance covered when he finally comes to a stop? (A) 22 m (B) 28 m (C) 36 m (D) 41 m

Sign Conventions a 13 m s-1 (f) (i) +ve s1 s2 25 m s1 = ut = 13 x = 9.1 m v2 = u2 + 2as 02 = (+132) + 2(-4.5)s2 s2 = 18.8 m Total distance s = s1 + s2 = = 27.9 m (B)

Effect of Air Resistance
v W U Fv (= kv) Terminal vel v W U Fv (= kv) Low vel U W Initially, v = 0 m s-1 U + Fv = W At terminal velocity: Recall ?

Projectile Motion A projectile motion can be considered as a combination of two independent components of motion: (i) horizontal motion with constant speed throughout, zero acceleration (since there is no acceleration horizontally), assuming there is no air resistance. more more (ii) vertical motion with uniform acceleration (for example due to gravity g). More (fire max range) More (monkey)

Projectile Motion Horizontally: u cos u cos u cos u u sin u cos 
The horizontal motion is a constant velocity motion. Hence the equation of motion is simply sx = (u cos) t

Projectile Motion Vertically: g u u sin  u cos
The vertical motion is a uniformly accelerated motion (acc = g). The equations of motion are given by the kinematics equations e.g. v = usin gt sy = (usin)t ½(g)t2 v2 = (usin) (g)sy

Projectile Motion Steps to success in projectile motion: Draw a good diagram with all the data shown. Draw the path of the motion of the body for better visualisation. Decide on your sign convention. Indicate your initial and final positions of the body on the path of the motion. Decide on the direction of analysis (vertical or horizontal). Use the appropriate kinematics equation.

Projectile Motion eg) A stone is projected horizontally with an initial velocity of 30 m s-1 above the horizontal from a cliff top which is 55 m above the sea level. What is the time taken for the stone to reach the sea? 55 m s 30 m s-1

which direction to analyse?
which equation to use? sign convention? Projectile Motion 55 m s 30 m s-1 (i) (f) Vertically: s = ut + ½ gt2 55 = 0 + ½ (9.81) t2 t = 3.34 s ( s is inadmissible)

Projectile Motion eg) A stone is projected vertically upwards with an initial velocity of 25 m s-1 above the ground. Determine the highest height reached by the stone. 25 m s-1 s

Projectile Motion (f) 25 m s-1 s which direction to analyse? which equation to use? sign convention? Vertically: v2 = u2 + 2gs (i) 0 = (-9.81)s s = 31.9 m

Projectile Motion eg) A stone is projected with an initial velocity of 50 m s-1 at angle of 28 above the horizontal as shown below. Calculate the maximum horizontal distance reached by the stone. s 50 m s-1 28

t = 4.79 s (t=0 is inadmissible)
Projectile Motion 50 cos 28o m s-1 50 sin 28o m s-1 50 m s-1 28 (f) (i) s vertically: s = ut + 1/2gt2 0 = (50 sin 28o)t – ½ (9.81)t2 t = 4.79 s (t=0 is inadmissible) horizontally: s = ut + 1/2gt2 = (50 cos 28o)4.79 – 0 = 211 m

Projectile Motion eg) A stone is projected with an initial velocity of 12 m s-1 at angle of 30 above the horizontal from a cliff top which is 75 m above the sea level. What is the time taken for the stone to reach the sea? (A) s (B) s (C) s (D) s 75 m s 12 m s-1 30 (B)

Projectile Motion 75 m s 12 m s-1 30 (i) 12 sin30o 12 cos30o 12 30o +ve (f) Vertically: Using t = 4.6 s or – 3.3 s (inadmissible)

Projectile Motion (b) Determine the position of the stone from the cliff when it reaches the sea. 12 m s-1 30 (i) +ve 75 m (f) s horizontally: Using s = ut = 12 cos30o x 4.6 = 47.8 m (A) m (B) m (C) m (D) m (A)

Projectile Motion Truck and ball Plane and package

The End

Take A Break! A clock chimes 5 times in 4 seconds. How many times will it chime in 10 seconds? 11 times. It chimes at zero and then once every second for 10 seconds.

Take A Break! What goes up, must come down.

Take A Break! Before it starts...
A man comes home from an exhausting day at work, plops down on the couch in front of the television, and tells his wife, "Get me a beer before it starts!" The wife sighs and gets him a beer. Ten minutes later, he says, "Get me another beer before it starts!" She looks cross, but fetches another beer and slams it down next to him. He finishes that beer and a few minutes later says, "Quick, get me another beer, it's going to start any minute!" The wife is furious. She yells at him "Is that all you're going to do tonight! Drink beer and sit in front of that TV! You're nothing but a lazy, drunken, fat slob, and furthermore..." The man sighs and says, "It's started..."

A rich man's son was kidnapped
A rich man's son was kidnapped. The ransom note told him to bring a valuable diamond to a phone booth in the middle of a public park. Plainclothes police officers surrounded the park, intending to follow the criminal or his messenger. The rich man arrived at the phone booth and followed instructions but the police were powerless to prevent the diamond from leaving the park and reaching the crafty villain. What did he do? Take A Break! This is a true story from Taiwan. When the rich man reached the phone booth he found a carrier pigeon in a cage. It had a message attached telling the man to put the diamond in a small bag which was around the pigeon's neck and to release the bird. When the man did this the police were powerless to follow the bird as it returned across the city to its owner.

Take A Break! After teaching his class all about roman numerals (X = 10, IX=9 and so on) the teacher asked his class to draw a single continuous line and turn IX into 6. The only stipulation the teacher made was that the pen could not be lifted from the paper until the line was complete. Draw an S in front of the IX and it spells SIX. No one said the line had to be straight :)

Take A Break! A babysitter came over one day to babysit 10 children. She decided to give them a snack. In a jar there were 10 cookies. She wants to give each one a cookie, but still keep one in the jar. How will she do it? (WITHOUT BREAKING ANY COOKIES-EACH CHILD HAS TO GET A WHOLE!) She hands the 10th child the jar with one cookie left in it.

Take A Break!

Take A Break! After teaching his class all about roman numerals (X = 10, IX=9 and so on) the teacher asked his class to draw a single continuous line and turn IX into 6. The only stipulation the teacher made was that the pen could not be lifted from the paper until the line was complete. Draw an S in front of the IX and it spells SIX. No one said the line had to be straight :)

Take A Break! 4 new fathers.
Four expectant fathers were in a Minneapolis hospital waiting room while their wives were in labor. The nurse arrived and proudly announced to the first man, "Congratulations, sir. You're the father of twins!" "What a coincidence! I work for the Minnesota Twins Baseball team!" Later the nurse returned and congratulated the second father on the birth of his triplets. "Wow! That's incredible! I work for the 3M Corporation." An hour later, the nurse returned to congratulate the third man on the birth of his quadruplets. Stunned, he barely could reply, "I don't believe it! I work for the Four Seasons Hotel!" After this, everyone turned to the fourth guy who had just fainted. The nurse rushed to his side. As he slowly gained consciousness, they could hear him mutter over and over, "I should never have taken that job at 7-Eleven. I should never have taken that job at 7-Eleven. I should never have taken that job...." Take A Break!

Take A Break! This is a true story. A white horse jumped over a tower and landed on a priest, who immediately disappeared from the landscape. Where did this take place? A chess board. The white horse = knight The tower = rook The priest = bishop

Take A Break! A man was drinking in a bar when he noticed this beautiful young lady sitting next to him. ''Hello there,'' says the man, ''and what is your name?'' ''Hello,'' giggles the woman, ''I'm Stacey. What's yours?'' ''I'm Jim.'' ''Jim, do you want to come over to my house tonight? I mean, right now??'' ''Sure!'' replies Jim, ''Let's go!'' So Stacey takes Jim to her house and takes him to her room. Jim sits down on the bed and notices a picture of a man on Stacey's desk. ''Stacey, I noticed the picture of a man on your desk,'' Jim says. ''Yes? And what about it?'' asks Stacey. ''Is it your brother?'' ''No, it isn't, Jim!'' Stacey giggles. Jim's eyes widen, suspecting that it might be Stacey's husband. When he finally asks, ''Is it your husband?'' Stacey giggles even more, ''No, silly!'' Jim was relieved. ''Then, it must be your boyfriend!'' Stacey giggles even more while nibbling on Jim's ear. She says, ''No, silly!!'' ''Then, who is it?'' Jim asks.

Take A Break! Stacey replies, ''That's me BEFORE my operation!!''

Take A Break! You have a 5 gallon bucket and a 3 gallon bucket,and a hose to fill them up,and you need to get 4 gallons. You have no means of measuring how many gallons are in each bucket (except knowing the buckets capacity) how can you be certain that you have 4 gallons? Answer Fill up the 3 gallon bucket and pour it in to the five gallon bucket(You now have 3 gallons in the 5 gallon bucket). Fill up the three gallon bucket again and pour it into the five gallon bucket until it is full.(You now have 5 gallons in the five gallon bucket and 1 gallon in the 3 gallon bucket.) Dump the water out of the 5 gallon bucket. Pour the one gallon from the 3 gallon bucket into the 5 gallon bucket. (You now have 1 gallon in the five gallon bucket) Fill up the 3 gallon bucket and pour it into the 5 gallon bucket. (You now have 4 gallons in the 5 gallon bucket.

Take A Break! Why is a river so rich? Because it has two banks!
What is in between you? The letter ‘O’ Which two days in a week starts with "T" other than tuesday and thursday? Today and tomorrow A women who works in a sweet shop has a measurement of , is 5'4" tall, wear size "6" shoes. What do you think she weighs? Sweets, what else?

Take A Break! What question can you never answer YES? Are you asleep?

Velocity-Time Graph eg) The figure below shows a velocity-time graph for a journey lasting 65 s. It has been divided up into six sections for each case of reference. 10 20 30 40 50 60 70 -5 v/m s-1 t/s A B C D E F

Velocity-Time Graph (a) Using information from the graph obtain: (i) the velocity 10 s after the start, velocity 10 s after the start = 20 m s-1 (ii) the acceleration in section A. acceleration in Section A = = 2.0 m s-2

Velocity-Time Graph (iii) the acceleration in section E, acceleration in Section E = = m s-2 (iv) the distance travelled in section B, Distance travelled in Section B = 20 x 15 = 300 m (v) the distance travelled in section C. Distance travelled in Section C = ½( )(10) = 250 m

(b) Sketch the corresponding displacement-time graph.
Velocity-Time Graph (b) Sketch the corresponding displacement-time graph. Distance from start A B C D E F linear linear quad quad linear quad t/s 10 20 30 40 50 60 70

H-I Body undergoing constant deceleration and comes to rest at I.
v t A B C D E F G H I Area 1 Area 2 Velocity-Time Graph E-F Stationary. F-G Body moving in opposite direction with magnitude increasing, or accelerating in opposite direction. G-H Constant velocity. H-I Body undergoing constant deceleration and comes to rest at I. Total distance moved = area 1 + area 2 Net displacement = area 1 – area 2

Sign Conventions eg) A ball is projected upwards from point A. Identify the sign of the displacement, velocity and acceleration during its flight, taking the upwards direction as positive. s v a 1 2 3 4 5 A 1 2 3 4 5 +ve + + -g + -g + - -g - -g - - -g

Sign Conventions eg) A man throws a ball vertically upwards with a velocity of 20 m s-1. Neglecting air resistance, find (i) the maximum height reached (f) At maximum height, v = 0 m s-1, Using v2 = u2 + 2as 02 = (+20)2 + 2(-9.81) s s = 20.4 m. (i) +ve

Sign Conventions (ii) the time taken for the ball to return to the man’s hand. using v = u + at 0 = (+20) + (-9.81)t t = 2.04 s Total time of flight = 2 x 2.04 = 4.08 s (i) (f) +ve

eg) A commando launches his parachute 20 s after jumping off a helicopter. Draw a graph to illustrate how his velocity vary with time from the moment he jumps of his helicopter till the moement he lands safely. v 20 A B C t

At A: As the commando falls through the air, he experiences a stronger and stronger viscous force. As a result, his acceleration gradually decreases from 9.8 m s-2 to zero. Before he opens his parachute, he is already moving at terminal velocity which has a large magnitude.

Fv (= kv) Fv (= kv) v W U Fv (= kv) Terminal vel U U v W W

At B: After 20 s he opens he parachute. This increases his cross-sectional area immediately and he experiences a much larger viscous force. At this moment, U + Fv > W. So he starts to decelerates until U + Fv = W. At C: When U + Fv = W is achieved, he reaches a new terminal velocity which is much lower than the previous one. This will allow him to land safely.

Projectile Motion eg) A bomber, flying at a horizontal speed of 360 km h-1 and at an altitude of 2 km above sea level, wishes to attack an enemy vessel. Calculate the angle of sight , at which the bomber should release its bomb so that it would most probably hit the vessel. 2 km 360 km h-1 x  Plane and package

Projectile Motion (i) +ve (f) Vertically: Using
2 km 360 km h-1 x  (i) +ve (f) Vertically: Using = 0 + 1/2(-9.81) t2 t = 20.2 s

Projectile Motion (i) +ve (f) Horizontally:  = 44.7o 360 km h-1 
x  (i) +ve (f) Horizontally:  = 44.7o

Required time taken = 16.3 – 0.88 = 15.4 s (i) (f)
Projectile Motion eg) A missile is fired vertically upward with an initial velocity of 84 m s-1 from a point level with the foot of a tower 70 m high. Calculate the time from when the missile is first level with the top of the tower until it is again level with the top of the tower. 84 m s-1 +ve Vertically: Using Required time taken = 16.3 – 0.88 = 15.4 s (i) (f) 70 m

Take A Break!

Take A Break! video

Take A Break! The marksman?

Take A Break! What question can you never answer YES? Are you asleep?

Take A Break! What goes up, must come down.

Take A Break! A man was drinking in a bar when he noticed this beautiful young lady sitting next to him. ''Hello there,'' says the man, ''and what is your name?'' ''Hello,'' giggles the woman, ''I'm Stacey. What's yours?'' ''I'm Jim.'' ''Jim, do you want to come over to my house tonight? I mean, right now??'' ''Sure!'' replies Jim, ''Let's go!'' So Stacey takes Jim to her house and takes him to her room. Jim sits down on the bed and notices a picture of a man on Stacey's desk. ''Stacey, I noticed the picture of a man on your desk,'' Jim says. ''Yes? And what about it?'' asks Stacey. ''Is it your brother?'' ''No, it isn't, Jim!'' Stacey giggles. Jim's eyes widen, suspecting that it might be Stacey's husband. When he finally asks, ''Is it your husband?'' Stacey giggles even more, ''No, silly!'' Jim was relieved. ''Then, it must be your boyfriend!'' Stacey giggles even more while nibbling on Jim's ear. She says, ''No, silly!!'' ''Then, who is it?'' Jim asks.

Studying the motion of bodies??? …….
The motivation behind …….

Take A Break! Stacey replies, ''That's me BEFORE my operation!!''

Take A Break! What about this??? Which one is false?

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