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Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 2 Motion in 1 Dimension.

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Presentation on theme: "Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 2 Motion in 1 Dimension."— Presentation transcript:

1 Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 2 Motion in 1 Dimension – Part 2

2 Physics 203 – College Physics I Department of Physics – The Citadel Today’s Topics & Announcements 1. Continuously Changing Velocity - Graphical 2. Acceleration - Graphical 3. Constant Acceleration Motion and Falling We will work a couple of problems similar to what you will do in the homework, and discuss problem solving approaches.

3 Physics 203 – College Physics I Department of Physics – The Citadel Thursday’s Assignment Read Ch. 3, sec. 1 – 4. The topic is vectors. Problem Set HW2 is open and due Thursday. It is mostly on Chapter 2, but includes some trigonometry practice for Chapter 3. We will review some trigonometry next time. A problem set on HW3 on Ch. 3 will be due the following Thursday. You may start it as soon as it is posted. A reading quiz is a possibility for any lecture. Always come prepared.

4 Physics 203 – College Physics I Department of Physics – The Citadel Non-Constant Velocity Estimate the velocity at t = 10 s. v = Δ x/ Δ t = 5 m / 17.5 s = 0.29 m/s

5 Physics 203 – College Physics I Department of Physics – The Citadel Non-Constant Velocity At what time is v = 0? v = 0 at the turning point, t = 37.5 s

6 Physics 203 – College Physics I Department of Physics – The Citadel Non-Constant Velocity Estimate the velocity at t = 30 s. v = Δx/Δt = 25 m / (37.5 – 16) s = 1.1 m/s

7 Physics 203 – College Physics I Department of Physics – The Citadel Velocity Graph Average acceleration is the change in velocity over a time interval. What is the average acceleration for the first 50 s? a = Δ v/ Δ t = (38 – 14) m/s 50 s = 0.48 m/s 2

8 Physics 203 – College Physics I Department of Physics – The Citadel Velocity Graph What is the average acceleration from 50 s to 100 s? a = Δ v/ Δ t = (0 – 38) m/s 50 s = – 0.76 m/s 2

9 Physics 203 – College Physics I Department of Physics – The Citadel Train A train car moves along a straight track. The graph shows that it (A)Speeds up all the time (B)Slows down all the time (C)Speeds up some of the time and slows down some of the time. (D)Has constant velocity. t x

10 Physics 203 – College Physics I Department of Physics – The Citadel Acceleration and Speed Can an object at rest have nonzero acceleration? A. Yes B. No Otherwise it would remain at rest. If an object has positive acceleration, does that mean its speed is increasing? A. Yes B. No If its velocity is negative, its speed is increasing.

11 Physics 203 – College Physics I Department of Physics – The Citadel Example If a car initially in reverse comes to rest, its acceleration is positive, but its speed is decreasing. t v v0v0 0

12 Physics 203 – College Physics I Department of Physics – The Citadel Constant Acceleration When the acceleration is constant, the position vs time graph forms a parabola. The velocity vs time graph is linear. The velocity is zero at the turning point of the motion. t (s) x (m) t (s) v (m/s)

13 Physics 203 – College Physics I Department of Physics – The Citadel Constant Acceleration Galileo dropped balls from the Tower of Pisa to show that the time to fall did not depend on the weight. The gravitational acceleration is g = 9.80 m/s 2 at the Earth’s surface. The motion forms a parabola, with the distance traveled given by y = ½ g t 2.

14 Physics 203 – College Physics I Department of Physics – The Citadel Dropping vs Throwing If you drop an object, and can neglect air resistance, it accelerates downward at 9.80 m/s 2. If you throw it downward, after you let go, it (A) accelerates at less than 9.8 m/s 2 (B) accelerates at 9.8 m/s 2 (C) accelerates at more than 9.8 m/s 2

15 Physics 203 – College Physics I Department of Physics – The Citadel Constant Acceleration Equations x = x 0 + v 0 t + ½ a t 2 (quadratic equation) v = v 0 + a t (linear equation) v 2 = v a Δ x ( Δ x = x – x 0 ) v = ½ (v 0 +v f ) (average velocity is in the middle)

16 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown Upward When you throw a ball straight up, which are true at the highest point? (A) The velocity and acceleration are both zero. (B) The velocity is nonzero and the acceleration is zero. (C) The acceleration is nonzero and the velocity is zero. (D) The velocity and acceleration are both nonzero.

17 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown Upward If a ball is thrown upward at 4.9 m/s at point A, how long does it take to return to be caught at point C? Let y = 0 at A or C. A → B: v = v 0 – gt = 0 t = v 0 /g = 4.9/9.8 s = 0.5 s. The time to fall is the same, so the ball takes 1.0 s to be caught.

18 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown Upward How high is the ball at point B, its high point? y = v 0 t – ½ gt 2 v 0 = 4.9 m/s, t = 0.5 s. y = (4.9 m/s)(0.5 s) + ½ (9.8 m/s 2 )(0.5 s) 2 = 3.7 m.

19 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown from Cliff A ball is thrown vertically upward at 5.0 m/s and lands at the base of the cliff 20 m below. (a) How fast is it falling just before it lands? (b) How long is it in the air?

20 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown from Cliff (a) Choice of axis: I’ll take y = 0 at the edge of the cliff, with positive y upward. (a) v f 2 = v a (y – y 0 ) = v gh = (5 m/s) 2 + 2(9.8 m/s 2 )(20 m) = 417 m 2 /s 2. v f = – 20.4 m/s ≈ – 20 m/s

21 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown from Cliff (b) t = (v f – v 0 )/a = (v 0 – v f )/g = (5.0 m/s – (– 20.4 m/s))/(9.8 m/s 2 ) = 2.59 m/s ≈ 2.6 s. Alternative: You could use the fact that y = y 0 + v 0 t + ½ at 2. – 20 m = 0 + (5.0 m/s) t – (4.90 m/s 2 ) t 2

22 Physics 203 – College Physics I Department of Physics – The Citadel Ball Thrown from Cliff 4.90 t 2 –5 t – 20 = 0 in seconds. Quadratic formula: If At 2 + Bt + C = 0, t = (5 ± 20.4) / 9.8 = 2.6 s or – 1.6 s.

23 Physics 203 – College Physics I Department of Physics – The Citadel Up vs Down A man on a cliff at the edge of a crater on the moon throws one ball straight up and another straight down at the same initial speed. Which hits the bottom of the crater at the higher speed? A.The one thrown up hits faster. B.The one thrown down hits faster. C.They hit at the same speed.


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