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Definition of Acceleration

The direction of accel- eration is same as direction of force. The acceleration is proportional to the magnitude of the force. Definition of Acceleration. An acceleration is the change in velocity per unit of time. (A vector quantity.)

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Definition of Acceleration

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  1. The direction of accel- eration is same as direction of force. • The acceleration is proportional to the magnitude of the force. Definition of Acceleration • An acceleration is the change in velocity per unit of time. (A vector quantity.) • A changeinvelocity 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:

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

  3. slope v2 Dv Dv v1 Dt Dt t1 t2 time Average and Instantaneous a

  4. Force + v1= +8 m/s 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? t = 4 s v2= +20 m/s 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.

  5. Force + t = 4 s v2 = +20 m/s v1= +8 m/s Step 5. Recall definition of average acceleration. Example 3 (Continued):What is average acceleration of car?

  6. + Force E 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.) vf= -5 m/s vo= +20 m/s Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. Step 3. Label given info with + and - signs.

  7. 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, vo=+20 m/s, east (+) Final velocity, vf = -5 m/s, west (-) The change in velocity, Dv = vf - v0 Dv = (-5 m/s) - (+20 m/s) = -25 m/s

  8. vf - vo tf - to Dv Dt aavg= = vo= +20 m/s vf= -5 m/s + Force Dv = (-5 m/s) - (+20 m/s) = -25 m/s E -25 m/s 5 s a = Example 4: (Continued) Acceleration is directed to left, west (same as F). a = - 5 m/s2

  9. C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E 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

  10. x = 0 C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E 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.

  11. C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E Signs for Acceleration What are the signs (+ or -) of acceleration at points B, C, and D? • At B, C, and D, a = -5 m/s, negative at all points. • The force is constant and always directed to left, so acceleration does not change.

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

  13. Formulas based on definitions: Derivedformulas: For constant acceleration only

  14. 0 0 Use of Initial Position x0in Problems. If you choose the origin of your x,y axes at the point of the initial position, you can set x0 = 0, simplifying these equations. The xoterm is very useful for studying problems involving motion of two bodies.

  15. Review of Symbols and Units • Displacement (x, xo); meters (m) • Velocity (vf, vo); meters per second (m/s) • Acceleration (a); meters per s2 (m/s2) • Time (t); seconds (s) Review sign convention for each symbol

  16. 2 m The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. -1 m -2 m The Signs of Displacement • Displacement is positive (+) or negative (-) based on LOCATION.

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

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

  19. 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,vo,a,t) Find: ____, _____ • Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

  20. +400 m/s v = 0 300 m vo + F X0 = 0 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? Step 1. Draw and label sketch. Step 2. Indicate + direction and F direction.

  21. +400 m/s v = 0 300 m vo + F X0 = 0 Example: (Cont.) Step 3. List given; find information with signs. Given:vo = +400 m/s v = 0 x = +300 m List t = ?, even though time was not asked for. Find:a = ?; t = ?

  22. x +400 ft/s v = 0 300 ft vo + 0 0 F X0 = 0 -vo2 2x -(400 m/s)2 2(300 m) a = = Continued . . . Step 4. Select equation that contains aand not t. 2a(x -xo) = v2 - vo2 Initial position and final velocity are zero. a = - 267 m/s2 Why is the acceleration negative? Because Force is in a negative direction!

  23. CONCLUSION OF Chapter 6 - Acceleration

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