August 31, 2001 Dr. Larry Dennis, FSU Department of Physics Physics 2053C – Fall 2001 Discussion of Chapter 2 Examples of 1-D Motion & Free Fall

Motion with Constant Acceleration x = x 0 + v 0 t + ½ a t 2 v = v 0 + a t v 2 = v a (x – x 0 ) x 0 is the initial position v 0 is the initial velocity

Sample Problem 1. How fast is it moving at the end of 5 seconds? 2. How far has it gone in the first 5 seconds? 3. What is it’s average speed for the first 5 seconds? 4. How far does it go between 5 and 10 seconds? 5. What is it’s average velocity over 10 seconds? A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed.

General Equations A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. x o = 0 m v o = 0 m/s a = 3 m/s 2 for t  5 s X = X o + V o t + ½ a t 2 X = t + 3/2 t 2 X = 3/2 t 2 V = V o + a t V = 3 t

Part 1 A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. V = V o + a t V = 3 t V = 3 m/s 2 * 5 s = 15 m/s 1. How fast is it moving at the end of 5 seconds?

Part 2 A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. X = X o + V o t + ½ a t 2 X = ½ 3 t 2 = 3 / 2 t 2 X = 1.5 m/s 2 * 5 2 s 2 = 37.5 m 2. How far has it gone in the first 5 seconds?

Part 3 A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. 3. What is it’s average speed for the first 5 seconds? V ave = (x 2 – x 1 ) / (t 2 – t 1 ) V ave = (37.5 – 0)m / (5 – 0)s = 7.25 m/s = 7.3 m/s

Part 4 A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. X = X o + V o t + ½ a t 2 X = 0 + (15 m/s) t + ½ * 0 * t 2 X = 15 m/s * 5 s = 75 m 4. How far does it go between 5 and 10 seconds?

Part 5 A police car starts from rest and accelerates at a constant rate of 3 m/s 2 for 5 seconds and then continues moving with constant speed. V ave = (x 2 – x 1 ) / (t 2 – t 1 ) X 2 = distance traveled in first 5 seconds + distance traveled in second 5 seconds. = 37.5 m + 75 m = m V ave = (112.5 – 0)m / (10 – 0)s = 11.3 m/s 5. What is his average velocity over 10 seconds?

Free Fall Acceleration due to gravity (abbreviated as “g”) downward constant 9.8 m/s 2 (we often approximate this at 10 m/s 2 ) X = X o + V o t + ½ a t 2 Need to be careful about directions: Either up or down can be chosen to be positive If up is positive, the acceleration is negative. If down is positive, the acceleration is positive.

Free Fall - Sample Problem A baseball is thrown upward at 40 m/s. 1. How high is at 2 s? 2. How fast is it moving at 2 s? 3. When does it reach it’s maximum height? 4. How high does it go? 5. When does it come back to it’s starting point? 6. How fast is it going when it reaches it’s starting point.

Free Fall – General Solution A baseball is thrown upward at 40 m/s. x o = 0 m v o = 40 m/s a = -10 m/s 2 X = X o + V o t + ½ a t 2 X = t - 5 t 2 X = 40 t – 5 t 2 V = V o + a t V = 40 – 10 t t(s) x(m)

Free Fall – Part 1 A baseball is thrown upward at 40 m/s. 1. How high is at 2 s? X = 40 t – 5 t 2 X = 40 m/s * m/s 2 * 2 2 s 2 X = 80 m – 20 m = 60 m

Free Fall – Part 2 A baseball is thrown upward at 40 m/s. 2. How fast is it moving at 2 s? V = V o + a t V = 40 – 10 t V = 40 m/s – 10 m/s 2 * 2 s V = 40 m/s – 20 m/s = 20 m/s

Free Fall – Part 3 A baseball is thrown upward at 40 m/s. 3. When does it reach it’s maximum height? It reaches it’s maximum height when V = 0. V = V o + a t V = 40 – 10 t 0 = 40 – 10 t  10 t = 40  t = (40 m/s) / (10 m/s 2 ) = 4 s.

Free Fall – Part 4 A baseball is thrown upward at 40 m/s. 4. How high does it go? Calculate it’s position at t = 4 s. X = X o + V o t + ½ a t 2 X = t - 5 t 2 X = 40 t – 5 t 2 X = 40 m/s * 4 s – 5 m/s 2 * 4 2 s 2 X = 160 m – 80 m = 80 m

Free Fall – Part 5 A baseball is thrown upward at 40 m/s. 5. When does it come back to it’s starting point? Find the times when X = 0. X = X o + V o t + ½ a t 2 X = 40 t – 5 t 2 0 = ( 40 – 5 t ) t t = 0 or 0 = ( 40 – 5 t )  40 = 5 t  t = 8 s. When an object is thrown upward and lands at the same height it takes just as long to go up as is does to come down.

Free Fall – Part 6 A baseball is thrown upward at 40 m/s. 6. How fast is it going when it reaches it’s starting point. Find the velocity at t = 8 s. V = V o + a t X = 40 – 10 t X = ( 40 m/s – 10 ms/s 2 * 8 s) X = 40 m/s – 80 m/s = - 40 m/s When an object is thrown upward and lands at the same height it’s speed when it reaches the bottom is the same as when it was thrown?

CAPA Starts Wednesday!

First Lab - Math Review ExercisesPage A: 1, 42 B: 22 C: 1, 32 D: 1, 42 E: 2, 54 F: set 1: 1, 3, 6, 10; set 2: 34-5; 5-6 G: 2, 4, 67 H: set 1: 1; set 2: 2, 38; 9 I: set 1: 2, 4; set 2: 2, , 13 set 3: 2; set 4: 1, 313, 15

Next Time Study Chapter 3. Start: Vectors. Motion in two dimensions. No class on Monday. Labs start on Tuesday. First CAPA set available in class on Wednesday. See me with any questions or comments. See you Wednesday.