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x 2. x 1. Motion Along a Straight Line. When the relevant behavior of an object can be adequately described by a single coordinate it is treated as a point particle . Motion in 1-Dimension Choose x-axis to lie along object’s motion Displacement = change in value of objects x-coordinate
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x2 x1 Motion Along a Straight Line • When the relevant behavior of an object can be adequately described by a single coordinate it is treated as a point particle. • Motion in 1-Dimension • Choose x-axis to lie along object’s motion • Displacement = change in value of objects x-coordinate • Average Velocity: • over time interval t1 to t2 Dx = x2 – x1
Instantaneous Velocity • Rate at which displacement changes with time http://phys23p.sl.psu.edu/simulations/1d_motion.ip • slope of graph of x versus t. • generally not equal to instantaneous velocity. • +/- slope for • +/- velocity (direction!) • motion away from/towards the origin
Average acceleration • A change in the (instantaneous) velocity with time • acceleration is not the same as velocity! • Instantaneous acceleration • Rate at which velocity changes with time
What can be said about displacement, velocity and acceleration at and between t1 and t2? x-t graph x t t1 t2 x-t graph x t t1 t2
What can be said about displacement, velocity and acceleration at and between t1 and t2? v-t graph v t t1 t2 v-t graph v t t1 t2
Motion with constant acceleration • a simplified model for many useful situations • a = aav • a = constant means v-t graph is a straight line • with the conventions: • velocity is v at time t, • velocity is v0 at time 0 :
at at at at Velocity and Acceleration final velocity = initial velocity + change in velocity v v v t v0 v v0 t v v t v0 v v0 t v
constant acceleration a with the conventions: • position is x, velocity is v at time t, • position is x0, velocity is v0 at time 0 :
Summary: For a particular problem, use an equation which only has one unknown
Example 2-4 A motorcyclist heading east is 5.00m past the city limits sign (taken to be the origin) at a speed of 15.0 m/s. His acceleration is 4.00 m/s2. • Find his position and velocity 2.00 s later. • Where is the motorcycle when his velocity is 25.0 m/s. • First identify known quantities, then identify appropriate equation to solve for desired unknown.
Example 2-5 A motorist traveling with a constant velocity of 14 m/s (about 34 mph) passes a school crossing, where the speed limit is 15 mph. Just as the motorist passes, a police officer on a motorcycle waiting at the crossing accelerates at a rate of 3.00 m/s2 in hot pursuit of the motorist. • How much time elapses before the officer catches up with the motorist? • What is the officer’s speed at this point? • How far beyond the crossing are the vehicles at this point?
Free fall: under the influence of gravity • Aristotle’s vs Galileo’s picture of motion • acceleration due to gravity (on the surface of the Earth) • air resistance can be ignored • acceleration is downward • if positive direction is taken to be upward, a = -g
Example 2-6 An object is dropped from the top of a very tall building. How far has it dropped and how fast is it moving after • 1.00s? • 2.00s? • 3.00s?
Example 2-7,8 An object is thrown straight upward with a speed of 15.0 m/s from the roof of a very tall building. On the way down, it just misses the railing. • Find the position and velocity of the object 1.00 and 4.00 seconds after being thrown. • Find the velocity when the object is 5.00 m above the railing. • Determine the maximum height reached, and the time at which it is reached. • The acceleration of the object when it is at its maximum height. • At what time is the object 5.00 m below the railing?