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 x2 x1 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

x-t graph What can be said about displacement, velocity and acceleration at and between t1 and t2? x t t1 t2 x-t graph x t t1 t2

v-t graph What can be said about displacement, velocity and acceleration at and between t1 and t2? 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 :

Velocity and Acceleration final velocity = initial velocity + change in velocity v v v at t v0 v at v0 t v v t v0 at v v0 t at 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 :

One final useful relation, take

Summary: For a particular problem, use an equation which only has one unknown

Example 2-4 A motorcyclist heading east is 5 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 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 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?