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Chapter 2: Motion Along a Straight Line

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1 Chapter 2: Motion Along a Straight Line
AP Physics Miss Wesley

2 Objectives In this chapter you will be able to have a mathematical description of motion For now, we don’t care what is causing the motion. For now, consider point-like objects (particle)

3 Definitions Position: the position of a particle can be specified by some number along the x-axis. Here x~3.7 m Displacement: The change in position of an object. x = x2-x1 Example: A particle moves from x1 = -2.0 m to x2 = 3.6 m. Find the displacement. x = x2-x1 = 3.6m – (-2m) = 5.6 m Total Displacement = 5.6 m -2 -1 1 2 3 4 5

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 time Vav = (x2-x1) (t2-t1) Speed: the average speed is the total distance traveled by an object in a certain amount of time. Note: sav ≠ vav sAv = total distance t x t Slope of this line = vav Vav = 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: Start 5.2 miles at 43 mph 1.2 miles Finish at Gas station

6 Velocity vs. Speed Example con’t
A) Find vav: x = displacement from start to finish = 4.0 miles Need to find t. tdriving = 5.2 miles/(43 mi/h) = h twalking = 0.5 h t = h h = h Vav = 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 miles Total 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 t0 = x/t = dx/dt This is the derivative of x with respect to t. Notations review: v = instantaneous velocity vav = v = average velocity

10 Acceleration When the instantaneous velocity is changing with time, then it is accelerating Average 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 = limt0 = v/t

12 Review Average Velocity: vav = x/t = (x2-x1) (t2-t1) Average Speed:
sav = total distance/ t Instantaneous Velocity: v = dx/dt = derivative of x with respect to t slope 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 velocities Instantaneous acceleration: A = dv/dt = derivative of v with respect to t OR – the slope of the tangent line on a v vs. t graph OR  a = d2x/dt2 = the second derivative of x with respect to t.

14 Brief Intro To Derivatives – Power Rule
Suppose- x= ctn Evaluate - dx / dt dx/dt =nc*t(n-1) Example 1: x=t2 dx/dt = 12t = v Example 2: x= -5t3 +6t dt/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+125 when t in seconds and y in meters. a) Find the velocity as a function of time b) 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 Increasing x x Speed Decreasing Speed Increasing x Speed Decreasing x

19 Motion with Constant Acceleration
Acceleration does not change with time SPECIAL 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. x xo vo

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 + at Graph will increase linearly because only one multiple of t. Slope equals acceleration. vo v t

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 acceleration v = vo + at x = xo + vot + ½at2 v2 = vo2 + 2a(x-xo)

24 The Acceleration of Gravity
An example of motion with constant acceleration Experiments show that ALL objects fall to the Earth with constant “free-fall” acceleration g = 9.81 m/s2 This means that heave objects fall at the same rate as light objects (ignoring air resistance)

25 Free Fall Motion We can use (1), (2), & (3) to describe free fall motion with a few changes Because yes, it does have constant acceleration y-axis is the direction of free-fall. It will point upward. a = -g because objects fall downward.

26 New Equations: v = vo – gt y = yo + vot - ½gt2 v2 = 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 ground Measure 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 2 A 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?

31 Example 2 con’t What is the max. height?

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/dt Suppose we are given v(t), how can we find the displacement (Δx) between ta and tb? USE INTEGRATION!

33 Velocity vs. Time Graph

34 Integration con’t We can make the equation exact by taking the limit as Δti  0 Δx = 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”


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