1 / 25

Should be : Wednesday, Sept 8

Should be : Wednesday, Sept 8. Announcements. You may attend any of the instructors’ office hours. List maintained on Blackboard under “Staff / Office Hours”. HW #1 due tonight at 11:59 p.m. HW #2 now available, due Thurs Sept 9 at 11:59 p.m.

papina
Download Presentation

Should be : Wednesday, Sept 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Should be: Wednesday, Sept 8

  2. Announcements • You may attend any of the instructors’ office hours. List maintained on Blackboard under “Staff / Office Hours”. • HW #1 due tonight at 11:59 p.m. • HW #2 now available, due Thurs Sept 9 at 11:59 p.m. • Solutions to HW #1 will be available on Blackboard (under “Solutions”) after the due date/time. • Quiz grades will be posted to Blackboard after your recitation instructor has graded the quizzes (also returned next week).

  3. Last Time: Motion in One Dimension Displacement, Velocity, Acceleration • Today: One-Dimensional Motion with Constant Acceleration Freely Falling Objects (under gravity)

  4. Constant/Uniform Acceleration • Constant/Uniform Acceleration: magnitude and direction of the acceleration does not change. • Constant acceleration is important, because it applies to many natural phenomena. • One such example is objects in “free fall” near the surface of the Earth. Neglecting air resistance, all objects (independent of mass) “fall” with the same downward acceleration !

  5. Constant Acceleration • Key Point: • Under constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the average acceleration over the entire time interval. This Means: velocity slope = a v at Let: ti = 0, vi = v0, tf = t, and vf = v v0 v0 time t (for constant a only)

  6. Constant Acceleration velocity velocity a < 0 a > 0 v v0 time time Δt Average velocity in any time interval Δt is : Just the average of the velocities at the start/finish of the time interval ! (for constant a only)

  7. Constant Acceleration Recall: The displacement Δx and average velocity v Set: If ti = 0 : (for constant a only) Using v = v0 + at : (for constant a only)

  8. Constant Acceleration velocity slope = a v For constant acceleration, the displacement Δx is just equal to the area under the velocity-vs-time graph ! at v0 v0 time t Recall: Area of Triangle = ½ (base) (height)

  9. Constant Acceleration Finally: Recall and Plugging in for t gives us: (for constant a only)

  10. “Equations of Motion”for Constant Acceleration Information Equation Velocity as a function of time Displacement as a function of time Velocity as a function of displacement • These are for motion along a single axis (x-axis) with acceleration alongthat direction. • If acceleration and motion along y-axis, just change x  y. • Assumes that at t = 0, v = v0. • Recall: Δx = x – x0

  11. Example A car starting from rest undergoes a constant acceleration of a = 2.0 m/s2. What is the velocity of the car after it has traveled 100 m? How much time does it take to travel the 100 m?

  12. Example: Problem 2.27 A car traveling east at 40.0 m/s (= 89.48 mph !) passes a trooper hiding at roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.50 s. What is the magnitude and direction of the car’s acceleration as it slows down? How far does the car travel during the 3.50-s time period? v0 = 40 m/s West East -x +x

  13. Example: Problem 2.29 A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final velocity of 2.80 m/s. Find the truck’s original speed. Find its acceleration.

  14. Freely Falling Objects • Key Point: • When air resistance is negligible, all objects falling under the influence of gravity near the Earth’s surface fall at the same constant acceleration. Aristotle (384 – 322 B.C.) Plato (428 – 348 B.C.) Heavier objects fall faster than lighter objects.

  15. Galileo Galilei (1564-1642) • Italian physicist and philosopher • Professor at the University of Padua • Invented the thermometer and the pendulum clock “Galileo, perhaps more than any other single person, was responsible for the birth of modern science.” – Stephen Hawking

  16. Galileo Galilei (1564-1642) • First to observe the heavens with a telescope • Defended the (disturbing) idea of Copernicus (1473-1543) that the Earth was not the center of the universe • Tried for heresy by Catholic Church (1992: Vatican found him not guilty) • Undertook series of experimental studies to describe the motion of bodies: mechanics

  17. Freely Falling Objects • What is a freely falling object ? • Any object moving freely under the influence of gravity alone, regardless of its initial motion. • Denote magnitude of free-fall acceleration as: • g = 9.80 m/s2(magnitude on the surface of the Earth) • If we: • Neglect air resistance • Assume acceleration doesn’t vary with altitude (over short vertical distances) “up” = +y Motion of freely falling object is the same as one-dimensional motion with constant acceleration !

  18. +y = “up” “Equations of Motion”for Freely Falling Objects Information Equation Velocity as a function of time Displacement as a function of time Velocity as a function of displacement Recall: Δy = y – y0

  19. Objects Rising Against Gravity If object’s initial (upward) velocity is v0 , how high will it rise? y y = 0

  20. Objects Rising Against Gravity If object’s initial (upward) velocity is v0 , how high will it rise? First, solve for the time to reach the highest point.  At highest point, v = 0 ! y Second, calculate the maximum height after solving for the time to reach the highest point. Alternative method … y = 0

  21. Example: Multiple Choice #10 (p. 48) A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground … True or False: The velocities of the two balls are equal. y v0 y = h v0 h y = 0

  22. Example: Multiple Choice #10 (p. 48) A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground … True or False: The speed of each ball is greater than v0. y v0 y = h v0 h y = 0

  23. Example: Multiple Choice #10 (p. 48) A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground … True or False: The acceleration of the green ball is greater than that of the red ball. y v0 y = h v0 h y = 0

  24. Example: Problem 2.50 A small mailbag is released from a helicopter that is descending steadily at 1.50 m/s. After 2.00 s: What is the speed of the mailbag? ; and How far is it below the helicopter ? How do (a) and (b) change if the helicopter is rising steadily at 1.50 m/s?

  25. Reading Assignment • Next class: 3.1 – 3.3

More Related