2. Definition of Acceleration  An acceleration is the change in velocity per unit of time. (A vector quantity.)  A change in velocity requires the application of a push or pull (force). A formal treatment of force and acceleration will be given later. For now, you should know that: The direction of accel- eration is same as direction of force. The acceleration is proportional to the magnitude of the force.

3. The Signs of Acceleration Acceleration is positive (+) or negative (-) based on the direction of force.Acceleration is positive (+) or negative (-) based on the direction of force. Choose + direction first. Then acceleration a will have the same sign as that of the force F — regardless of the direction of velocity. F F + a (-) a (+)

4. Average and Instantaneous a vv tt v2v2 v1v1 t2t2 t1t1 vv tt time slope

5. Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s. What is average acceleration? Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right). Step 3. Label given info with + and - signs. Step 4. Indicate direction of force F. + v 1 = +8 m/s t = 4 s v 2 = +20 m/s Force

6. Example 3 (Continued): What is average acceleration of car? Step 5. Recall definition of average acceleration. + v 1 = +8 m/s t = 4 s v 2 = +20 m/s Force

7. Example 4: A wagon moving east at 20 m/s encounters a very strong head-wind, causing it to change directions. After 5 s, it is traveling west at 5 m/s. What is the average acceleration? (Be careful of signs.) Step 1. Draw a rough sketch. + Force Step 2. Choose the eastward direction as positive. v o = +20 m/s v f = -5 m/s Step 3. Label given info with + and - signs. E

8. Example 4 (Cont.): Wagon moving east at 20 m/s encounters a head-wind, causing it to change directions. Five seconds later, it is traveling west at 5 m/s. What is the average acceleration? Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity,  v = v f - v 0  v = (-5 m/s) - (+20 m/s) = -25 m/s Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity,  v = v f - v 0  v = (-5 m/s) - (+20 m/s) = -25 m/s

9. Example 4: (Continued) a avg = a avg = = vvttvvtt v f - v o t f - t o t f - t o a = a = a = a = -25 m/s 5 s 5 s a = - 5 m/s 2 Acceleration is directed to left, west (same as F). + Force v o = +20 m/s v f = -5 m/s E  v = (-5 m/s) - (+20 m/s) = -25 m/s

10. Signs for Displacement Time t = 0 at point A. What are the signs (+ or -) of displacement at B, C, and D? At B, x is positive, right of origin At C, x is positive, right of origin At D, x is negative, left of origin + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D

11. Signs for Velocity What are the signs (+ or -) of velocity at points B, C, and D?  At B, v is zero - no sign needed.  At C, v is positive on way out and negative on the way back.  At D, v is negative, moving to left. + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D x = 0

12. What are the signs (+ or -) of acceleration at points B, C, and D?   The force is constant and always directed to left, so acceleration does not change.  At B, C, and D, a = -5 m/s, negative at all points. Signs for Acceleration + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D

13. Constant Acceleration Acceleration: Setting t o = 0 and solving for v, we have: Final velocity = initial velocity + change in velocity

14. Formulas based on definitions: Derivedformulas Derived formulas: For constant acceleration only

15. Use of Initial Position x 0 in Problems. If you choose the origin of your x,y axes at the point of the initial position, you can set x 0 = 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies. 00

16. Review of Symbols and Units (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v f, v o ); meters per second (m/s)Velocity (v f, v o ); meters per second (m/s) ( a ); meters per s 2 (m/s 2 )Acceleration ( a ); meters per s 2 (m/s 2 ) Time (t); seconds (s)Time (t); seconds (s) (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v f, v o ); meters per second (m/s)Velocity (v f, v o ); meters per second (m/s) ( a ); meters per s 2 (m/s 2 )Acceleration ( a ); meters per s 2 (m/s 2 ) Time (t); seconds (s)Time (t); seconds (s) Review sign convention for each symbol

17. The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION.Displacement is positive (+) or negative (-) based on LOCATION. The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. 2 m -1 m -2 m

18. The Signs of Velocity Velocity is positive (+) or negative (-) based on direction of motion.Velocity is positive (+) or negative (-) based on direction of motion. First choose + direction; then velocity v is positive if motion is with that + direction, and negative if it is against that positive direction. + - - + +

19. Acceleration Produced by Force Acceleration is (+) or (-) based on direction of force (NOT based on v).Acceleration is (+) or (-) based on direction of force (NOT based on v). A push or pull (force) is necessary to change velocity, thus the sign of a is same as sign of F. F a (-) a (-) F a (+) More will be said later on the relationship between F and a.

20. Problem Solving Strategy:  Draw and label sketch of problem.  Indicate + direction and force direction.  List givens and state what is to be found. Given: ____, _____, _____ (x,v,v o, a,t) Find: ____, _____   Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

21. Example 6: A airplane flying initially at 400 m/s lands on a carrier deck and stops in a distance of 300 m. What is the acceleration? 300 m +400 m/s v o v = 0 + F Step 1. Draw and label sketch. F Step 2. Indicate + direction and F direction. X 0 = 0

22. Example: (Cont.) 300 m +400 m/s v o v = 0 + F Step 3. Step 3. List given; find information with signs. Given: v o = +400 m/s v = 0 v = 0 x = +300 m Find: a = ?; t = ? List t = ?, even though time was not asked for. X 0 = 0

23. Step 4. a t Step 4. Select equation that contains a and not t. 300 ft +400 ft/s v o v = 0 + F x 2 a (x -x o ) = v 2 - v o 2 00 a = = -v o 2 2x -(400 m/s) 2 2(300 m) a = - 267 m/s 2 a = - 267 m/s 2 Why is the acceleration negative? Continued... Initial position and final velocity are zero. X 0 = 0 Because Force is in a negative direction!

24. CONCLUSION OF Chapter 6 - Acceleration