A Riddle to get us started: A boat has a ladder that has six rungs. Each rung is 20 centimeters apart. The bottom rung is one foot from the water. The tide rises at 12 inches every 15 minutes. High Tide peaks in one hour. An inch is made up of 2.54 cm. When the tide is at its highest, how many rungs are under water?

Representing Motion

Motion We are looking to classify and compare an object in motion. Three “rules” we will follow: –The motion is in a straight line –The cause of the motion is ignored (coming soon!) –The objects considered is a particle (not for long!) Particles and particle like objects move uniformly –Ex. A sled going down a hill –ANTI Ex. A ball rolling down a hill

Position The location of the particle in space. Needs a mathematical description to be useful. We assign a number to represent the particles position on a coordinate grid. –There needs to be a zero point to reference –The positions to the left are negative –The positions to the right are positive

Distance Distance: The total path length when moving from one location to another –Scalar Has only magnitude (i.e. size) and unit (NO DIRECTION) Other scalars: mass, time, energy

Displacement Displacement: the straight-line distance between two points, along with the direction from the starting point to the finish point –Vector: Has a magnitude, unit and direction Other vectors: acceleration, momentum, etc. Ex. 23 m/s [W] or -23 m/s

Displacement cont’d Position: x (m) Displacement: Δx = x 2 - x 1 or x f – x i +x→ ← -x +x→ ← -x

Displacement vs. Distance +x→ ← -x +x→ ← -x

Speed Speed: Rate which distance is traveled –Average Speed ( ): The distance, d, traveled over the total time of the trip –Instantaneous speed: speed at a particular instant

Speed cont’d Instantaneous or Average Speed?

Speed cont’d Example 1: A car is moving at a constant speed. If the car traveled a distance of 60 meters in 4 seconds: a) Find the speed of the car At the 60 meter mark, the car suddenly slows down to rest over 3.5 seconds and covers another 15 meters in doing so. b) Find the average speed of the car over the course of the entire problem

Velocity Velocity: how fast something is moving and in which direction Direction of velocity is determined by the direction of the displacement

Instantaneous Velocity Instantaneous velocity: How fast something is moving at a particular time (w/ a direction) Defined later as the slope of a position-time (x vs t) graph

Riddle A man wishes to marry a wealthy kings daughter. The wealthy king, hoping to make a fool of the man, gives him a test. If the man passes, the king explains, he will be married to the daughter. The man is blindfolded and taken outside. There are 20 statues in a line. All the statues are black except for one which is white. The blindfolded man must find the white statue to marry the daughter. How does he find it?

Graphing Motion Carl Lewis ‘88Usain Bolt ‘ m1.89 s1.85 s m2.96 s2.87 s m3.90 s3.78 s m4.79 s4.65. s m5.65 s5.50 s m6.48 s6.32 s m7.33 s7.14 s m8.18 s7.96 s m9.04 s8.79 s m9.92 s9.69 s

Bolt ‘08 vs. Lewis ‘88

Position (x) vs Time Graphs 30 m 40 m50 m60 m20 m10 m0 m 3 s 4 s5 s6 s2 s1 s0 s X(m) t(s) 0 m 0 s 6 s 60 m x(m)t(s)

Avg Velocity vs Instantaneous Velocity On a Position vs. Time graph: –Avg Velocity is the displacement divided by time interval over which it occurred –Instantaneous velocity is the slope of a line at a given point If the slope is constant along a line segment Avg. Velocity = Inst. Velocity If the slope is changing v inst = slope of a line tangent at a given point. X(m) t(s) 5

1.position at t = 7 s 2.distance from 2 to 4 s 3.distance from 0 to 5 s 4.displacement from 0 to 4 s 5.displacement from 0 to 3 s 6.Time interval where speed is changing 7.speed at t = 2.5 s 8.instantaneous speed at t = 4 s 9.average velocity from 3 to 7 s 10.average speed from 0 to 8 s

Graphing Summary up to now x(m) t(s) Straight line means NO acceleration x(m) t(s) Curved line means changing slope which means changing VELOCITY

Interpreting a VvT Graph v(m/s) t(s) +, unchanging velocity v(m/s) t(s) +, increasing velocity v(m/s) t(s) +, decreasing velocity v(m/s) t(s) -, increasing velocity v(m/s) t(s) -, decreasing velocity

Interpreting a VvT Graph v(m/s) t(s) v(m/s) t(s) v(m/s) t(s) v(m/s) t(s) If the line is in the positive (+) portion of the graph the object is moving forward (i.e., + direction) If the line is in the negative (-) portion of the graph the object is moving backward (i.e., - direction)

Interpreting a VvT Graph v(m/s) t(s) v(m/s) t(s) If the line passes from the one region (+ to – or – to +) to another, the object changes direction

Time (s) v (m/s) Interpreting a Velocity vs. Time Graph The area under the curve is the objects displacement.

Interpreting a Velocity vs. Time Graph The area under the curve is the objects displacement. Time (s) v (m/s)

Interpreting a Velocity vs. Time Graph The area under the curve is the objects displacement. Time (s) v (m/s)

What is the displacement from 0s to 2s? What is the displacement from 6s to 7 s? What is the displacement from 8s to 10s?

Interpreting a VvT Graph The slope of a line on a VvT graph indicates acceleration (unit: m/s 2 ). Time (s) v (m/s)

Acceleration Acceleration: The time rate of change of velocity Vector (has magnitude and direction) Unit: m/s 2 The slope of a line on a velocity vs. time graph

Visualizing Acceleration

Practice Problem The United States Bowling Congress conducted a study on ideal bowling ball speed. It was found that a bowling ball should leave the hand going 9.4 m/s. If the ball goes from being at rest to 9.4 m/s in 1.5 seconds what is the acceleration of the ball?

Practice Problem A man starts from rest and is accelerated at a rate of m/s 2 over a time interval of.75s. What was his final velocity?

Acceleration Expressed in g’s –When accelerations are large we express them as a multiple of “g” It is the acceleration due to gravity near the surface of the Earth

Constant Acceleration This is a special case that tends to simplify things. Constant, or mostly constant, acceleration occurs all the time. –Car starting from rest when a light turns green –Car braking at a light when a light turns red There are a set of equations that are used to describe this motion.

Kinematic Equations

Constant Acceleration Problem A car starts from rest and accelerates uniformly to 23 m/s in 8 seconds. What distance did the car cover in this time?

Graphical Look at Motion: displacement – time curve The slope of the curve is the velocity The curved line indicates the velocity is changing –Therefore, there is an acceleration

Graphical Look at Motion: velocity – time curve The slope gives the acceleration The straight line indicates a constant acceleration

The zero slope indicates a constant acceleration Graphical Look at Motion: acceleration – time curve

Test Graphical Interpretations Match a given velocity graph with the corresponding acceleration graph

Free Fall Acceleration This is a case of constant acceleration that occurs vertically. All things fall to the Earth with the same acceleration –In the absence of air resistance, all things fall to the Earth with the same acceleration: –This is invariant of the objects dimensions, density, weight etc. When using the kinematic equations we use –a y = -g = m/s 2

Free Fall – an object dropped Initial velocity is zero Let up be positive Use the kinematic equations –Generally use y instead of x since vertical Acceleration is –a y = -g = m/s 2 v o = 0 a = -g

Free Fall – an object thrown downward a y = -g = m/s 2 Initial velocity  0 –With upward being positive, initial velocity will be negative v o ≠ 0 a = -g

Free Fall -- object thrown upward Initial velocity is upward, so positive The instantaneous velocity at the maximum height is zero a y = -g = m/s 2 everywhere in the motion v = 0 v o ≠ 0 a = -g

Thrown upward, cont. The motion may be symmetric –Then t up = t down –Then v = -v o The motion may not be symmetric –Break the motion into various parts Generally up and down

Free Fall Example Initial velocity at A is upward (+) and acceleration is -g (-9.8 m/s 2 ) At B, the velocity is 0 and the acceleration is -g (-9.8 m/s 2 ) At C, the velocity has the same magnitude as at A, but is in the opposite direction The displacement is –50.0 m (it ends up 50.0 m below its starting point)

Vertical motion sample problem A ball is thrown upward with an initial velocity of 20 m/s. –What is the max height the ball will reach? –What will the velocity of the ball be half way to the maximum height? –What will the velocity of the ball be half way down to the hand? –What is the total time the ball is in the air?