Acceleration To calculate acceleration we use the following

Acceleration • To calculate acceleration, we use the following formula: • This can also be written as

Example Exercises from Notes • Accelerating Plane 2 can also be • An acceleration of +4. 35 m/s -2. written as +4. 35 m/s/s or +4. 35 m s This means that every second, the velocity changes by +4. 35 m/s

Acceleration § Which one of these accelerations is the largest? • 60 mi/hr-s • 60 mi/hr-min • 60 mi/hr-hr

Ex. Exercises from Notes, cont. • Car Slowing Down • Notice that in computing Δ v, you always subtract final from initial: v-v 0 • Question: An object moving at -25 m/s with an acceleration of +5 m/s/s. What is its velocity after 7 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 the average acceleration? + Force t=4 s v 1 = +8 m/s v 2 = +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.

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

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. ) + v f = -5 m/s E Force v o = +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.

Example 4 (Cont. ): Wagon moving east at 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, Dv = v f - v 0 Dv = (-5 m/s) - (+20 m/s) = -25 m/s 20

+ Example 4: (Continued) v f = -5 m/s aavg = Dv Dt a=-5 m/s 2 E v o = +20 m/s Force Dv = (-5 m/s) - (+20 m/s) = -25 m/s = vf - vo tf - t o a= -25 m/s 5 s Acceleration is directed to left, west (same as F).

+ Signs for Displacement D v f = -25 m/s A v o = +20 m/s C E Force a = - 5 m/s B 2 Time t = 0 at point A. What are the signs (+ or -) of displacement moving from? A to B B to C C to D

+ Signs for Velocity D v f = -25 m/s x=0 A v o = +20 m/s C E Force a = - 5 m/s B 2 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.

Signs for Acceleration + D v f = -25 m/s Force C A v o = +20 m/s B a = - 5 m/s 2 What are the signs (+ or -) of acceleration at points B, C, D? D , a = -5 m/s, § At B, and C, and negative at all points. § The force is constant and always directed to left, so acceleration does not change.

Distance and Acceleration • A car starting from rest accelerates at 2 m/s 2. Which of the following is true? A. The car covers 2 m of distance every second. B. The car covers 2 m more distance than in the previous second (2 m, then 4 m, then 6 m, etc) C. The car covers more and more distance each second (2 m, then 4 m, then 8 m,

Equations for Constant Acceleration To find the distance an object covers while accelerating, use this equation: (2) Notice that, if a =0 and you start from the origin, this equation becomes more recognizable: 0 0

Eqns for Motion at Constant Acceleration We can combine equations (1) & (2) so as to eliminate t : (3) We now have all the equations we need to solve constant-acceleration problems. (1) (2) (3)

Example 5: A ball 5 m from the bottom of an incline is traveling initially at 8 m/s. Four seconds later, it is traveling down the incline at 2 m/s. How far is it from the bottom at that instant? + d vo vf F -2 m/s t = 4 Careful 8 m/s s vo + v f 8 m/s + (-2 m/s) d = do + t =5 m+ (4 s) 2 2 5 m

(Continued ) + F d vo vf -2 m/s t = 4 8 m/s s 8 m/s + (-2 m/s) d=5 m+ (4 s) 2 5 m d=5 m+ 8 m/s - 2 m/s 2 (4 s) x = 17 m

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

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

Problem Solving Strategy: § Draw and label sketch of problem. § Set up coordinate system (which way is + and where is the origin) § List givens and state what is to be found. Given: ____, _____ ( d, v, v o , a, t ) Find: equation ____, _____ § Select containing one and not the other of the unknown quantities, and solve for the unknown.

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

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

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