What about this??? Which one is false?

Aim & Throw where????

Assessment Objectives: (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.

Assessment Objectives: (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 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: Definition:

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

Velocity 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). is not always equal to. Note: Definition:

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 What is the average speed of this car for this journey? A 12 m s -1 B 13.3 m s -1 C 40 m s -1 D 48 m s -1  A B = total distance / time

Velocity (Quick check!) Then same car now travels back from B to A with the following speeds: 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 D 96 m s -1 C 40 m s -1 B 48 m s -1  A B = 480 x 2 / 20

Velocity (Quick check!) What is the average velocity of this car for the whole journey? A 0 m s -1 D 96 m s -1 C 24 m s -1 B 48 m s -1  A B = displacement / time=0/20

Acceleration 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: Definition: 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? (c)a change in direction only e.g. circular motion. (b) a change in both magnitude and direction. (a)a change in magnitude e.g. body speeding up or slowing down.

Uniformly accelerated motion Refers to motion where the velocity of a body is changing at a steady rate. (constant acceleration) a toy car sliding down a slope a ball being thrown in the air. Eg) 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 s =  v dt v =  a dt v = ds / dt (gradient) a = dv / dt (gradient) s t s v t v a t s v

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

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

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?

Graph 1 Displacement-Time Graph

Graph 1: (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 = =

Graph 2 Displacement-Time Graph

Graph 2: (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.

Displacement-Time Graph Graph 2: (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 t s Motion under constant acceleration s =  v dt v =  a dt v = ds / dt a = dv / dt

Examples of Displacement-Time Graphs t s Motion under constant deceleration

Examples of Displacement-Time Graphs s t Motion under constant velocity

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

Velocity-Time Graph (Quick check!) 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

Velocity-Time Graph Deductions: 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. (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. v = u + at v 2 =u 2 + 2as Kinematics equations: Note: They are only valid for cases of uniform acceleration in a straight line. more

Sign Conventions (Quick check!) 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

v 2 = u 2 + 2as Sign Conventions 0 2 = (+13 2 ) + 2(-4.5)s 2 s 2 = 18.8 m Total distance s = s 1 + s 2 = = 27.9 m 13 m s m a s1s1 s2s2 (i) (f) +ve s 1 = ut = 13 x 0.70 = 9.1 m (B)

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

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. (ii) vertical motion with uniform acceleration (for example due to gravity g). More (fire max range) More (monkey) more

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

Projectile Motion Vertically:  u u cos  u sin  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 s y = (usin  )t - ½(g)t 2 v 2 = ( usin  ) 2 - 2(g)s y g

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. Use the appropriate kinematics equation. Decide on the direction of analysis (vertical or horizontal).

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

Projectile Motion 55 m s 30 m s -1 (i) (f) Vertically: s = ut + ½ gt 2 55 = 0 + ½ (9.81) t 2 t = 3.34 s ( s is inadmissible) which direction to analyse? which equation to use? sign convention?

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 25 m s -1 s (i) (f) which direction to analyse? which equation to use? sign convention? Vertically: v 2 = u 2 + 2gs 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 

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

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? 75 m s 12 m s  (A) 3.0 s(B) 4.6 s (C) 6.0 s(D) 8.5 s (B)

Projectile Motion Vertically: Using t = 4.6 s or – 3.3 s (inadmissible) 75 m s 12 m s  (i) (f) +ve 12 sin30 o 12 cos30 o o

Projectile Motion (b) Determine the position of the stone from the cliff when it reaches the sea. horizontally: Using s = ut = 12 cos30 o x 4.6 = 47.8 m 75 m s 12 m s  (i) (f) +ve (A) 47.8 m(B) 56.8 m (C) 70.2 m(D) 90.0 m (A)