DAY 1 Motion A Puzzler You ride your

DAY 1

Motion • A Puzzler – You ride your bike from Ossining to NYC, 30 miles away at 15 mph. How fast must you return to Ossining to average 30 mph?

Motion • A Puzzler – You ride your bike from Ossining to NYC, 30 miles away at 15 mph. How fast must you return to Ossining to average 30 mph? • Ans – It is impossible. You would have to return in 0 hours. Why? • Q- Determine the general formula for average velocity in terms of v 1 and v 2

Motion • V ave = 2 v 1 v 2 /(v 1 + v 2) • This is only true if the distance for trip 1 and 2 are the same.

Motion along a straight line • 1) We will only discuss straight line motion • 2) No causes of motion yet • 3) discuss all objects like particles • 4) Choose the center of mass as a point – like an elephant?

Displacement • Choice is + or – • Displacement, Dx = x 2 – x 1 • Displacement vs Distance • Ex) x 1 = 9 m x 2 = -6 m

Average Velocity • vave = total distance / total time • This really doesn’t tell you about the details of the trip. • More importantly you need to determine the instantaneous speed

Velocity • vave = Dx/Dt = x 2 – x 1/t 2 – t 1 • Its the slope of x vs t • But in order to get the instantaneous velocity, you must make the interval smaller and smaller, Dt approaches zero.

Calculus • v = lim as t approaches 0 of Dx/Dt = dx/dt in calculus words • dee x, dee t In calculus terms, the instant velocity is the rate at which an objects position is changing with respect to time at a given instant.

dx/dt = v • If we know the location of an object as a function of time, we can determine its speed. x = 5 t 2 dx/dt = v = d(5 t 2)/dt = 10 t • ex) x = 3 t 2 + 5 • • ex) x = t 2 - 8 t + 16

dx/dt = v • If we know the location of an object as a function of time, we can determine its speed. x = 5 t 2 dx/dt = v = d(5 t 2)/dt = 10 t • ex) x = 3 t 2 + 5 dx/dt = 6 t • • ex) x = t 2 - 8 t + 16 dx/dt = 2 t + 8

Acceleration • Acceleration, a, is the rate of change of velocity

Acceleration • Acceleration, a is the rate of change of velocity • a = Dv/Dt • = v 2 – v 1/(t 2 – t 1) • a = dv/dt = d 2 x/d 2 t the second derivative units are m/s/s

Below is an x vs t graph, draw the v vs t and a vs t graphs

• When v is constant, acceleration is zero graph of v vs t slope Dv/Dt • *** your body reacts to acceleration - not velocity. •

Given an x vs t graph, one can calculate the slope at every point in order to determine the v vs t graph and then the a vs t graph. Sketch these graphs on your dry erase boards.

Determine the v vs t and a vs t from this graph

Introduction to Integration • The area under v vs t is the displacement. This process is called the antiderivative or integral between two points. V in m/s t in s

Introduction to Integration • d = (vo + vf)/2 * t V in m/s t in s

Introduction to Integration • d = (vo + vf)/2 * t • or d = ½ at 2 + vot

Constant Acceleration • It is a special case when the acceleration is constant. • draw d vs t, v vs t, a vs t and compare use a = -9. 8 m/s/s

Equations for Constant a • v = vo + at • vave = (v + vo)/2 • x - xo = vave t • Dx = ½ at 2 + vot