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an ice skater coasting along a car screeching to a halt a ball dropped from the hand a feather floating in the wind a shell fired from a canon Motion – many examples surround us Of all these, simplest to analyze: ice skater coasting

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Useful Physical Quantities to Describe Motion Position x(t), as a function of time - describes the position of an object at any time. - a vector quantity (w/ magnitude, direction) - in metric units of meters Velocity v(t) - describes how position x(t) changes in time. - a vector quantity measured in units of meters/second v(t) = x/ t - magnitude of velocity is called speed Acceleration a(t) - describes how velocity v(t) changes in time. - a vector quantity measured in units of m/s 2 a(t) = v/ t

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Different Types of Motion 1.Uniform Motion (constant velocity, a = 0) speed v = x/ t speed is the magnitude of the velocity vector (a vector is a quantity w/ both magnitude and direction) Examples: a parked car skater coasting on a rink a car moving up a hill at steady speed and direction

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What happens when nothing is pushing her along? Why does she continue to coast? Coasting on Ice Skates reduce friction, making property of inertia more pronounced Mass of skater (measured in kg) is a measure of inertia (resistance to a change in motion). Anything with mass will continue in its current state of linear motion Newtons 1 st Law of Motion: An object that is not subject to any outside forces moves at constant velocity, covering equal distances in equal times, along a straight-line path.

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Everyday Demonstrations of Newtons First Law (Law of Inertia) Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator. The head of a hammer can be tightened onto the wooden handle by banging the bottom of the handle against a hard surface. To dislodge ketchup from the bottom of a ketchup bottle, you shake it. Headrests are placed in cars to prevent whiplash injuries during rear-end collisions.

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2. Uniformly Accelerated Motion (a 0) a = v/ t = (v-v o )/t …or v = v o + at x = x o + v o t + (1/2) a t 2 These two equations help you predict motion of uniformly accelerated objects. But why will an object accelerate in the first place, and change its velocity ? Newtons 2 nd Law of MotionF = m a A force F exerted on an object is equal to the product of the objects mass and its acceleration. The acceleration is in the same direction as the applied force.

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Examples of uniformly accelerated motion: your car when you drive it in a straight line with the accelerator pressed constantly a ball falling under the influence of gravity (freely-falling objects) Question: A car moving to the right begins to apply its brakes constantly. Where is the velocity pointing ? Where is the acceleration pointing ? v(t) a(t) Answer: v(t) pointing to the right, a(t) pointing to the left, opposite to the motion, causing the car to decelerate. Now: Lets take a closer look at the case of a ball dropped from rest.

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How does a dropped ball fall? (Describe a(t), v(t), x(t)) We use v = v o + at and x = x o + v o t + (1/2) a t 2 where a = constant, = -g = - 9.8 m/s2, downwards We can simplify matters by choosing x 0 (the initial position = 0); Result: v = v o – gt and x = - (1/2)gt 2 We can use these to simulate a ball dropping, finding its x and v at different times: t x(t)v(t) a(t) 0 secs 0 m0 m/s- 9.8 m/s 2 1 secs - 4.9 m - 9.8 m/s- 9.8 m/s 2 2 secs- 19.6 m -19.6 m/s- 9.8 m/s 2 3 secs- 44.1 m - 29.4 m/s- 9.8 m/s2

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Free-fall or falling - a special type of motion in which the only force acting upon an object is gravity Neglecting air friction, all objects will fall with the same acceleration, regardless of their mass. Why ? mouse Weight (force of gravity) acceleration

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A Closer Look: Big Rock vs Little Rock, which falls faster? Assuming no air friction. both fall at same rate

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Some Exercises 1.Bungee-Jumping Amusement How long should the rope be for a 5 – second free fall experience ? Solution: We know initial velocity v o = 0 and elapsed time t = 5 s How to find final v ? v = v o –g t = 0 – (10 m/s 2 )(5 s) = - 49 m, or 49 meters down 2. You throw a ball up with initial speed of 10 m/s. a. How high does it climb ? We know v o = 10m/s and v = 0 v = v o –g t or 0 = 10 m/s – (10m/s 2 ) t or t = 1 sec But x = v o t – (1/2)gt 2 = x = (10m/s)(1s) – (1/2)(-10m/s 2 )(1s) 2 or x = 5 meters How fast does it come back at you? (discussed in class)

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