2. Acceleration Is a change in velocity over time.Is a change in velocity over time. Is usually described as “speeding up” or “slowing down”, but ……Is usually described as “speeding up” or “slowing down”, but …… If an objects direction changes its direction, it is accelerating even if it is speed is not changing.If an objects direction changes its direction, it is accelerating even if it is speed is not changing. Is a vector!Is a vector! If acceleration and velocity are in same direction, the object speeds up. Opposite directions, the object slows down.If acceleration and velocity are in same direction, the object speeds up. Opposite directions, the object slows down.

3. Definitions Average velocity: Average acceleration:

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

5. Review of Symbols and Units (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v, v o ); meters per second (m/s)Velocity (v, 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) Values with an o subscript means it is the initial value. Values without it are final.Values with an o subscript means it is the initial value. Values without it are final. (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v, v o ); meters per second (m/s)Velocity (v, 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) Values with an o subscript means it is the initial value. Values without it are final.Values with an o subscript means it is the initial value. Values without it are final.

6. 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 0 0

7. Problem Solving Strategy:  Draw and label sketch of problem.  Indicate + direction.  List givens and state what is to be found. Given: ____, _____, _____ (x,v,v o, a,t) Unkowns: ____, _____   select Equation containing one and not the other of the unknown quantities, and Solve for the unknown.

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

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

10. 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 ft/s) 2 2(300 ft) a = - 267 ft/s 2 a = - 267 ft/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!

11. Acceleration Due to Gravity Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Earth Wg

12. Gravitational Acceleration In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: a = g = 9.8 m/s 2 or approximately 10 m/s 2 Directed downward (usually negative). Directed downward (usually negative).

13. Sign Convention: A Ball Thrown Vertically Upward Velocity is positive (+) or negative (-) based on direction of motion.Velocity is positive (+) or negative (-) based on direction of motion. Displacement is positive (+) or negative (-) based on LOCATION. Displacement is positive (+) or negative (-) based on LOCATION. Release Point UP = + Tippens Acceleration is (+) or (-) based on direction of force (weight).Acceleration is (+) or (-) based on direction of force (weight). y = 0 y = + y = 0 Negative y = - Negative v = + v = 0 v = - Negative v= - Negative a = -

14. Same Problem Solving Strategy Except a = g:  Draw and label sketch of problem.  Indicate + direction and force direction.  List givens and state what is to be found. Given: ____, _____, a = - 9.8 m/s 2 Find: ____, _____   Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

15. Example 7: A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s? Step 1. Draw and label a sketch. a = g + v o = +30 m/s Step 2. Indicate + direction and force direction. Step 3. Given/find info. a =-9.8 m/s 2 t = 2, 4, 7 s v o = + 30 m/s y = ? v = ?

16. Finding Displacement: a = g + v o = 30 m/s 0 y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2 Substitution of t = 2, 4, and 7 s will give the following values: y = 40.4 m; y = 41.6 m; y = -30.1 m Step 4. Select equation that contains y and not v.

17. Finding Velocity: Step 5. Find v from equation that contains v and not x: Substitute t = 2, 4, and 7 s: v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s a = g + v o = 30 m/s

18. Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity v f is zero. a = g + v o = +96 ft/s To find y max we substitute t = 3.00 s into the general equation for displacement. y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2

19. Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2 a = g + v o =+30 m/s t = 3.00 s y = 90.0 m – 45.0 m Omitting units, we obtain: y max = 45.0 m