2. DAY 1

5. 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?

6. 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

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

8. 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?

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

10. Average Velocity v ave = 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

11. Velocity v ave =  x/  t = 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,  t approaches zero.

12. Calculus v = lim as t approaches 0 of  x/  t = 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.

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

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

16. Acceleration Acceleration, a, is the rate of change of velocity

17. Acceleration Acceleration, a is the rate of change of velocity a =  v/  t = 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

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

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

20. 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.

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

24. 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

25. Introduction to Integration d = (v o + v f )/2 * t V in m/s t in s

26. Introduction to Integration d = (v o + v f )/2 * t or d = ½ at 2 + v o t

27. 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

28. Equations for Constant a v = v o + at v ave = (v + v o )/2 x - x o = v ave t  x = ½ at 2 + v o t