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

Example Exercises from Notes Accelerating Plane An acceleration of +4.35 m/s2 can also be written as +4.35 m/s/s or +4.35 m s-2. 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 Δv, you always subtract final from initial: v-v0 Question: An object moving at -25 m/s with an acceleration of +5 m/s/s. What is its velocity after 7 s?

+ Force t = 4 s v1 = +8 m/s v2 = +20 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 4s. What is the average acceleration? + Force t = 4 s v1 = +8 m/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.

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

+ E Force vf = -5 m/s vo = +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.) + Force E 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.

Choose the eastward direction as positive. 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

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

+ Signs for Displacement E Force C D B A vf = -25 m/s vo = +20 m/s a = - 5 m/s2 A B C D 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 E Force vo = +20 m/s vf = -25 m/s a = - 5 m/s2 A B C D x = 0 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 + Force vo = +20 m/s vf = -25 m/s a = - 5 m/s2 A B C D 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.

Distance and Acceleration A car starting from rest accelerates at 2 m/s2. Which of the following is true? The car covers 2 m of distance every second. The car covers 2 m more distance than in the previous second (2m, then 4m, then 6m, etc) The car covers more and more distance each second (2m, then 4m, then 8m, etc.)

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

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

+ d 8 m/s -2 m/s t = 4 s vo vf F d = do + t vo + vf 2 = 5 m + (4 s) 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? 5 m d 8 m/s -2 m/s t = 4 s vo vf F + Careful d = do + t vo + vf 2 = 5 m + (4 s) 8 m/s + (-2 m/s) 2

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

Use of Initial Position d0 in Problems If you choose the origin of your x,y axes at the point of the initial position, you can set d0 = 0, simplifying these equations. The do term is very useful for studying problems involving motion of two bodies.

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

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

Step 2. Indicate + direction and F direction. 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 vo v = 0 F X0 = 0 + Step 1. Draw and label sketch. Step 2. Indicate + direction and F direction.

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

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