Presentation on theme: "Chapter 2: Motion Along a Straight Line"— Presentation transcript:
1 Chapter 2: Motion Along a Straight Line AP PhysicsMiss Wesley
2 ObjectivesIn this chapter you will be able to have a mathematical description of motionFor now, we don’t care what is causing the motion.For now, consider point-like objects (particle)
3 DefinitionsPosition: the position of a particle can be specified by some number along the x-axis.Here x~3.7 mDisplacement: The change in position of an object.x = x2-x1Example: A particle moves from x1 = -2.0 m to x2 = 3.6 m. Find the displacement.x = x2-x1 = 3.6m – (-2m) = 5.6 mTotal Displacement = 5.6 m-2-112345
4 Position vs. Time Graphs A convenient way to depict the motion of a particle.Tells you the position of the particle at each instant in time.Velocity: vav = x/t x = change in position (displacement)t = change in timeVav = (x2-x1)(t2-t1)Speed: the average speed is the total distance traveled by an object in a certain amount of time.Note: sav ≠ vavsAv = total distancetxtSlope of this line = vavVav = slope of the line drawn between (t1, x1) and (t2, x2)
5 Velocity vs. Speed Example You drive down a road for 5.2 miles at 43 mph. You run out of gas and walk back to the gas station 1.2 miles away in 30 minutes.A) What is vav for the trip?B) What is the average speed for the trip?The first step is to draw a picture:Start5.2 miles at 43 mph1.2 milesFinish atGas station
6 Velocity vs. Speed Example con’t A) Find vav:x = displacement from start to finish = 4.0 milesNeed to find t.tdriving = 5.2 miles/(43 mi/h) = htwalking = 0.5 ht = h h = hVav = 4.0 mi/ h = 6.44 mi/h
7 Velocity vs. Speed Example con’t B) Find average speed:Total distance = 5.2 miles miles = 6.4 milesTotal time = h (from A)sav= 6.4 miles/ h= mi/h
8 Instantaneous Velocity How do we define the velocity of a particle at a single instant?When 2 points get close enough, the line connecting them becomes a tangent line.
9 Instantaneous Velocity con’t Instantaneous Velocity: slope of the tangent line to the x vs. t curve at a particular instant.v= instantaneous velocity = lim t0 = x/t = dx/dtThis is the derivative of x with respect to t.Notations review:v = instantaneous velocityvav = v = average velocity
10 AccelerationWhen the instantaneous velocity is changing with time, then it is acceleratingAverage Acceleration:aav = a = v/t = (v2-v1)/(t2-t1)All of the velocities are instantaneous velocities.This is the slope of the line connecting(t1, v1) and (t2,v2) on a velocity vs. time graph
11 Instantaneous Acceleration Instantaneous acceleration is the slope of a tangent line to a v vs. t curve at a particular instant.a = instantaneous acceleration = limt0 = v/t
12 Review Average Velocity: vav = x/t = (x2-x1) (t2-t1) Average Speed: sav = total distance/ tInstantaneous Velocity:v = dx/dt = derivative of x with respect to tslope of the tangent line on an x vs. t graph
13 Review con’t Average acceleration: Instantaneous acceleration: Aav = v/t = (v2-v1)/(t2-t1)The velocities are the instantaneous velocitiesInstantaneous acceleration:A = dv/dt = derivative of v with respect to tOR – the slope of the tangent line on a v vs. t graphOR a = d2x/dt2 = the second derivative of x with respect to t.
14 Brief Intro To Derivatives – Power Rule Suppose-x= ctnEvaluate - dx / dtdx/dt =nc*t(n-1)Example 1: x=t2dx/dt = 12t = vExample 2: x= -5t3 +6tdt/dt = v = -15t2+6
15 Brief Intro to Derivatives Example 3 – Suppose you know the height of a ball as a function of time.y(t)= -5(t-5) 2+125when t in seconds and y in meters.a) Find the velocity as a function of timeb) Find the acceleration as a function of time.
16 Example 3 – Derivatives – Chain Rule a) Find the velocity as function of time y(t)= -5(t-5) y(t)= -5(t2 -10t+25)+125 y(t)= -5t2+50t y’(t)= -10t+50= v(t) OR y’(t)=-10(t-5) = -10t+50=v(t)
17 Example – Second Derivative b) Find acceleration as a function of time. Recall v’(t)= a(t) y’(t)= v(t) = -10t+50 v’(t)=-10=a(t)
18 Check Point - Acceleration Find sign (+, -) of acceleration if….Speed IncreasingxxSpeed DecreasingSpeed IncreasingxSpeed Decreasingx
19 Motion with Constant Acceleration Acceleration does not change with timeSPECIAL CASE!It occurs often in nature (free fall)Consider: a particle which moves along the x-axis with constant acceleration a.Suppose at time t = 0s its initial velocity is vo, and its initial position is xo.xxovo
20 Motion with Constant Acceleration Using the previous situation, find the velocity at some time t.By definition:a = Δv/Δt = (v-vo)/(t-0)v = vo + atGraph will increase linearly because only one multiple of t.Slope equals acceleration.vovt
21 Motion with Constant Acceleration Find the position at some later time.
22 Motion with Constant Acceleration Another handy equation:Square both sides of equation 1
23 Review – Special Case of Motion with Constant Acceleration Before you use these formulas, you MUST make sure that the object has constant accelerationv = vo + atx = xo + vot + ½at2v2 = vo2 + 2a(x-xo)
24 The Acceleration of Gravity An example of motion with constant accelerationExperiments show that ALL objects fall to the Earth with constant “free-fall” accelerationg = 9.81 m/s2This means that heave objects fall at the same rate as light objects (ignoring air resistance)
25 Free Fall MotionWe can use (1), (2), & (3) to describe free fall motion with a few changesBecause yes, it does have constant accelerationy-axis is the direction of free-fall. It will point upward.a = -g because objects fall downward.
26 New Equations:v = vo – gty = yo + vot - ½gt2v2 = vo2 - 2g(y-yo)
27 Example: A ball is released from rest from a height h. How long does it take to hit the ground?
28 DEMO! Choose a location in the room from which to drop a ball. Measure the height, and determine the theoretical value for how long it should take the ball to hit the groundMeasure how long it actually takes the ball to hit the ground.Calculate the percent error between the measured time and the actual time
29 Example: A ball is released from rest from a height h. What is the ball’s velocity when it hits the ground? (the instant before when it actually hits the ground the velocity will be zero)
30 Example 2A pitcher can throw a 100 mph fast ball. If he throws the ball straight up, how long does it take to reach the highest point?
32 Integration The inverse operation of taking the derivative Recall: Given x(t), we can easily find v(t)V(t) is equal to dx/dtSuppose we are given v(t), how can we find the displacement (Δx) between ta and tb?USE INTEGRATION!
34 Integration con’tWe can make the equation exact by taking the limit as Δti 0Δx = lim Δti 0 Σi vi Δti = ∫tatb vdt“The itegral of vdt between ta and tb”Another interpretation:Δx = “area under the v vs. t curve”