Chapter 2: Motion Along a Straight Line - PowerPoint PPT Presentation

chapter 2 motion along a straight line n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 2: Motion Along a Straight Line PowerPoint Presentation
Download Presentation
Chapter 2: Motion Along a Straight Line

play fullscreen
1 / 35
Chapter 2: Motion Along a Straight Line
897 Views
Download Presentation
darin
Download Presentation

Chapter 2: Motion Along a Straight Line

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  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. -2 -1 0 1 2 3 4 5 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

  4. x Slope of this line = vav t 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 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: 5.2 miles at 43 mph Start 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) = 0.1209 h • twalking = 0.5 h • t = 0.1209 h + 0.5 h = 0.6209 h • Vav = 4.0 mi/0.6209 h = 6.44 mi/h

  7. Velocity vs. Speed Example con’t • B) Find average speed: • Total distance = 5.2 miles + 1.2 miles = 6.4 miles • Total time = 0.6209 h (from A) • sav= 6.4 miles/0.6209 h = 10.31 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: 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) 2+125 y(t)= -5(t2 -10t+25)+125 y(t)= -5t2+50t y’(t)= -10t+50= v(t) OR y(t)= -5(t-5) 2+125 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. Speed Increasing x Speed Decreasing x Speed Increasing x Speed Decreasing x Check Point - Acceleration • Find sign (+, -) of acceleration if….

  19. vo x xo 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.

  20. 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. v vo t Motion with Constant Acceleration

  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 0Σi vi Δti = ∫tatb vdt • “The itegral of vdt between ta and tb” • Another interpretation: Δx = “area under the v vs. t curve”