# Unit 2 - crsd.org

Unit 2 The Kinematics Equations (1D Equations of Motion) Day #1 Introduction to the Equation of Motion What are the 5 key equations we have learned thus far? A) x = xf xi finding a change in displacement B) v = vf vi C) s = d/t finding a change in velocity finding speed (distance/time)

D) v = x/t finding velocity (displacement/time) E) a = v/t finding acceleration (change in velocity change in time) There are 4 new equations for motion. Well start with a = v/t Substitute in v = vf vi now a = v -v The formula is f i t Rearrange to get at = vf - vi

Remember slope of the line = a = vf-vi t Example #1 A car starting from rest accelerates uniformly to a speed of 75 m/s in 12 seconds. What is the cars acceleration? First, identify the known values, then plug and chug Vi = 0 Vf = 75 m/ s t = 12 s a = ? *always

make a list a = vf-vi or at = vf - vi t = 75 m/s 0 12s =+6.25 m/s2 Example #2 A spacecraft traveling at 1025 m/s is uniformly accelerated at a rate of 105 m/s2 for 9 seconds. What is the final velocity of the spacecraft? First, identify the known values, rearrange the formula, then plug and

chug a = v -v or at = v f i f V = 1025 i m/s - vi Vf = ?

t t = 9 s Vf= Vi + at a = 105 m/ s2 = 1025 m/s + 9s(105m/ 2 The second new formula Recall: area under the curve on a v/t chart gives us displacement = x (or d) Consider a random, uniform accl object. Use the area of a trapezoid 1 2 area h(b1 b2 )

x 12 t (v1 v2 ) Example #3 A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop? Vi = 500 m/s Vf = 0 t = ? a = ? x = 15 m x =t(vf+vi) 15 m = (t)(0+500 m/hr) 30 m = 500m/hr(t) t = 0.06 hr

= 216 sec Example #3 A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop? What was its acceleration? Vi = 500 m/hr Vf = 0 t = 0.03 hr or 108 s (previous slide) a = ? x = 15 m a = vf-vi

t = 0-500m/hr 0.06 hr =-8333.33 m/hr2 Or -0.0006 m/s2 HW In book, Complete pg 53,55, & 58 odds Day #2 RE-Introduction to the equation of motion a v f vi t

v f vi at . area under curve = x 1 2 area Atriangle Arectangle bh BH x t (v f vi ) t (vi ) 1 2 1 2 2 x at v1t

2. a 2 sm2 vi 8 ms x 12 at 2 vi t 100 (2)t 8t 1 2 2 0 t 2 8t 100

You have to use the quadratic formula to solve the problem. x 100m The first answer doesnt make sense because you cant have a negative time. t 14.77sec, 6.77sec a = vf-vi

(rearrange) t v f vi at Substitute our newest formula into the derivation below v v at 2 f i 2 2

2 2 2 x at vi t 1 2 2 v f vi 2vi at a 2t 2 v f vi 2a (vi t 12 at 2 ) 2 2

v f vi 2a(x) Make a decision as to which direction is POSITIVE & which is NEGATIVE List what you have (using COMPATIBLE units & using proper SIGNS) choose your equation Solve the equation Make sure your answer makes sense (both in MAGNITUDE & DIRECTION) 3. vi 40 ms 2 v f 20 ms 2 v f vi 2ax

20 2 40 2 2a(50) a 12 sm2 x 12 t (vi v f ) 50 12 t(40 20) t 1.67sec x 50m 4.

vi 0 ms v f 60mph 26.82 ms t 3.5sec v f vi at 26.82 0 a(3.5) a 7.66 sm2 x0 1 12 at 2 vi t 12 (7.66)(1) 2 0(1) 3.83m x0 2 12 at 2 vi t 12 (7.66)(2) 2 0(2) 15.32m d0 1 3.83m

d1 2 15.32m 3.83m 11.49m Review Solve Solve the following: You roll a ball up an incline at a speed of 4.5 m/s. After 5 sec, the ball is on the way down travelling at a new speed of 1.5 m/s. Find the balls acceleration. (Sketch) -1.5-4.5 -6 -1.2 m/s2 5 5 rookie mistake. 1.5-4.5 Find the time to the peak. -1.2 = 0-4.5 -1.2 = -4.5

=3.75 sec t t How far up does it go? 0 = (4.5)2 + 2(-1.2)d =8.44 m Day #3 Free Fall Free Fall Free fall problems are acceleration problems where we know the acceleration. For all free fall problems (at the surface of the earth), a = 9.8 m/s2 (down) that the object speeds up by 9.8m/s2 after every second it falls

When solving: Make a chart Use arrows next to variables to show the direction of the vectors Fill in known variables Solve for unknowns Free Fall Problems Initial Velocity (Vi) will be 0, so our formulas change slightly. No need to memorize these.. 2 x at vi t 1 2

2 2 v f vi 2a (x) a = vf-vi t x at 1 2 2 2 v f 2a (x) a = vf t

9.8 m/s 2 0 m/s 9.8 m/s 19.6 m/s 29.4 m/s 39.2 m/s 49 m/s Free Fall Problems 1. A ball is dropped from the top of a bridge, and it takes 8 seconds to hit the water. A. How tall was the bridge?

B. How fast was the ball going just before it hits the water? a. height? So find x. b. a = (vf-vi)/t 9.8 x = a vi t +1/2(a)(t )2 9.8= vf/8 m/s2 2 = (9.8)(8) = 78.4 m/s 0 m = vi 313.6 downward vf x

*You could8ssolve part B first and then use the equation t x=1/2t(v i+vf) vi = 0 a 9.8m /s2 vi 0 a = -9.8 m/s2 x = -20 m 2

2 v f vi 2ax vf x 20m t 2 v f 0 2 2( 9.8)( 20) v f 392 v f 19.8 ms Because the diver is falling DOWNWARD,

we can also use a negative sign If all objects accelerate at the same rate when falling near the earths surface, why do some objects actually hit the ground faster than others? (assuming they are dropped at the same time from the same height)? AIR RESISTANCE!! Some objects are narrower (like pencils) and therefore the air resists their falling less. Some things are wider (like frisbees) and the air provides more resistance and slows down their fall. NOTICE HOW THE MASS OF THE OBJECTS DOES NOT MATTER! 5.

y v t Sometimes the variable y is used instead of x to simply show that the object is moving vertically t Notice that the object speeds up Notice that the acceleration is CONSTANT and NEGATIVE the

ENTIRE way Day #4 Throw-ups, Come-downs, Throw- downs Free Fall Problem Throw up A ball is thrown up from the ground In this type of problem, the initial velocity is always up. Also, the primary type of problem will ask what is the maximum height, which means vf = 0 m/s. 2. A ball is thrown up from the ground at an initial speed of 20 m/s. A. what is the maximum height the ball reaches? B. How long before it gets to the max altitude? A. x =? B. t =? Vf2=vi2 +2a x a=(vf-vi)/ t

-9.8m/ 2 0=2o a s2 +2(-9.8) x -9.8=(0-20) t x = 20.4m t=2.04s 20 m/s vi *You 0 can solve for B first then use the formula vf x =1/2t(vi+vf) x Problem #2

2 UP +, DOWN a = -9.8 m/s2 v2 = 0 (at the top) a) vi = 20 m/s a = -9.8 m/s2 vf = 0 m/s vf 2 2 1

2 vi 2ax 2 0 20 2( 9.8)x x 20.41m b) v f vi at 0 20 ( 9.8)t t 2.04 sec c) x 12 at 2 vi t

To solve for t, 11 12 ( 9.8)t 2 20t use the t= 0.66s and 3.42 sec quadratic (going up) (going down) formula Throw up/Come down Assume: Up+, Down A =-9.8m/s2 3 scenarios 3. 1.

v2= 0 (at the top) Well start with scenario #3. V3=-V1 2. Scenario 3 object is thrown up and caught at the same height 2 a) vf = 0 m/s a = -9.8 m/s2 x = 30 m 2 b )

a = -9.8 m/s2 vi= 24.25 m/s 2 v f vi 2ax 2 2 v f vi at vf = 0 m/s 0 vi 2( 9.8)(30) m vi 588 24.25 s

3 0 24.25 ( 9.8)t t 2.47 sec 1 t total 2(2.47sec) 4.94 sec c) Due to the fact that the ball is thrown and caught at the same height v 3 v1 24.25 ms Scenario 2 and part d of our problem 2 vi= 24.25 m/

s 3 1 4 a = -9.8 m/s2 x = -1 m (because down is -) vf2=vi2 +2ax vf2=24.252 +2(-9.8) (-1) 24.65 m/s down Or-24.65m/s What if we were solving for the time it was when the object hits the ground?

vi= 24.25 m/ sa = -9.8 m/ s2 x = -1 m (because down is -) x 12 at 2 vi t 1 12 ( 9.8)t 2 24.25t t .04 sec, 4.99 sec To solve for t, use the quadratic formula Only positive times make sense Day #5 Chase Problems

Chase Problems A chase problem is a scenario where two objects are involved, and they have the same position at some later time. They can Both start at the same place and same time Start at different places but at the same time Both start at the same place but at different times Start at different places at different times B or D but be moving in different directions The Key Equation x=1/2at2+vit x=xf-xi

Substitute the second formula into the first, creating: x2=1/2at2+vit+xi This formula will give you the final position for both objects. Since you dont know that final position, set the two formulas equal to each other. Chase Problem #1 Timmy is running at a constant speed of 6m/s. He sees Susie running 50m in front of him. She is moving at a constant speed of 4 m/s, in the same direction. How long will it take for Timmy to catch Susie? Notice that while Timmy and Susie have different starting

positions (xiT=0m while xiS=50m), they have the same final position (xf) after Timmy catches Susie. . . X xiT xiS x2 Tim = Susie /2at2+vit+xi = 1/2at2+vit+xi (0)t2+6t+0 = 1/2(0)t2+4t=50 6t = 4t=50 t = 25sec 1

Chase Problem #2 A continuation of the last problem. Timmy passes Susie, running at his constant speed of 6m/s. Susie decides to pick up the pace, very gradually. She begins to accelerate at a constant rate of 0.1m/s2. How long will it take her to catch Timmy? Notice that Timmy and Susie have the same starting position (x iT=0m xiS=0m) and ending position (xf) after Susie catches Timmy. . x Really, you could it=xis x2 have made x the initial position

anything you Tim = Susie wanted they 1/ at2+v t+x = 1/ at2+v t+x 2 i i 2 i i cancel out. I decided to choose (0)t2+6t+0 = 1/2(0.1)t2+4t+0 0. 6t=0.05t2+4t 6=0.05t+4 t=40sec

Chase Problem #3 A man drops a penny off the top of a 100m tall building. + 0 - Exactly 1 second later another man throws a nickel downward from the same place as the first man. What is the minimum speed the nickel must be thrown at in order to catch the penny? Start with the time it takes the penny to hit the bottom using the formula: x=1/2at2+vit -100=1/2(-9.8)t2+0t t= 4.52 sec penny

= nickel (1 sec later) 1/ at2+v t+x = 1/ at2+v t+x 2 i i 2 i i (-9.8)(4.52)2+0(4.52)+0 = 1/2(-9.8)(3.52)2+vi(3.52)+0 Vi= -11.19 m/s (neg. because its going down) Chase Problem #4 really hard In a strange, yet exciting, crash-test-dummy crash, two cars start by facing each other 1000m apart on a straight road. The first car accelerates from rest with a constant accel of 4m/s2. The second car accel, at a rate of 8 m/s2

for 5 sec but then settles into a constant speed. Find the elapsed time before these two cars collide. Notice that the dummies have different starting positions (xiA=0, xiB=1000m), but they have the same final position (x 2) when they crash. Start by finding the final velocity of car B by using the following formula: a = vf-vi -8=(vf-0)/5 so vf =-40m/s t 1 /2at2+vit+xi =

1 /2at2+vit+xi Test Review Step #1 Write Down What You Have (Look for Key Words) Coming to a stop Starting from rest Coasting Maximum Height Dropped vf = 0 vi = 0

vi = vf = constant vf = 0 a = -9.8 m/s2 Slowing Down a = - __ Braking Speeding up Accelerating from rest a = + __ Step #3 Solve the Equation

Step #4 Make sure your answer makes sense Helpful Tip #1 Choose your Key Points in every problemand do so wisely. Vertical Problems 1 Throwdowns Free-fall vi = 0

2 1 vi 0 2 Throw up 2 vf = 0 (at top) 1 Vertical Problems Throw up / Come Down

(throw and catch at same height) 2 vf = 0 (at top) v1 =-v3 1 3 Throw up / Come Down (throw and catch at different heights) 2 Use x ( 9.8)t vi t and solve quadratically for t 1 2

2 2 x13 = 3 1 1 x13 = 3 Helpful Tip #2 Assign positive and negative to different directions.

Helpful Tip #3 When solving a quadratic equation, do so with minimal effort. Solving a quadratic equation Choice B Choice A 2 b b 4ac 2a Factoring Unlikely on a physics problem

Choice C 2nd Trace Zero (on graphing calculator) 1 2 2 x at v1t Can be used at constant speeds (a=0) or when accelerating. Awesome Dude! Chase

Problems Since the two objects (A and B) end up at the same position by the end of the chase, use x 2A x 2B 1 2 2 1 2 2 at A v1A t A x1A at B v1B t B x1B

But what if The objects start at different places? x 2A x 2B 1 2 2 1 2 2 at A v1A t A x1A at B v1B t B x1B Its already accounted for

here and here But what if The objects start at different TIMES? x 2A x 2B 2 1 1 2 A 1A A 1A 2 Youll need to use an extra equation relating the two times. Plug this new

equation into the long equation above. at v t x 2 at B v1B t B x1B Example: tA = tB + 1 Helpful Tip #6 It is always important to remember that when something is thrown up or down (or simply falls), the acceleration at ALL times is constant.

+ - The acceleration of the ball at EVERY point on this red path (When its rising up, when its stopped, when its falling down) is always -9.8 m/s2. An object thrown up has a constant acceleration at ALL times. .the acceleration due to gravity x Objects rise and fall in the

same amount of time (assuming no parachute ) t v t Constant slope = constant accel. TONIGHTS HW Complete Review Worksheet

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