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**Unformatted text preview: **homework 03 – VENNES, ROSS – Due: Sep 21 2007, 1:00 am 1 Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 7 . 6 m / s. Also, it has an acceleration in the direction parallel to the walls of 2 . 6 m / s 2 . It hits the opposite wall at the same height. 18 . 4 m 2 . 6m / s 2 7 . 6 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 9 . 86832 m / s (tolerance ± 1 %). Explanation: Let : d = 18 . 4 m , v x = 7 . 6 m / s , a = 2 . 6 m / s 2 , Basic Concepts Kinematics equations v = v o + g t s = s o + v o t + 1 2 g t 2 d a 6 . 29474m / s 7 . 6 m / s 9 . 8 6 8 3 2 m / s 5 . 3 6 6 6 ◦ 7 . 61994 m The horizontal motion will carry the parti- cle to the opposite wall, so d = v x t f and t f = d v x = (18 . 4 m) (7 . 6 m / s) = 2 . 42105 s . is the time for the particle to reach the oppo- site wall. Horizontally, the particle reaches the maxi- mum parallel distance when it hits the oppo- site wall at the time of t = d v x , so the final parallel velocity v y is v y = a t = a d v x = (2 . 6 m / s 2 ) (18 . 4 m) (7 . 6 m / s) = 6 . 29474 m / s . The velocities act at right angles to each other, so the resultant velocity is v f = radicalBig v 2 x + v 2 y = radicalBig (7 . 6 m / s) 2 + (6 . 29474 m / s) 2 = 9 . 86832 m / s . Question 2, chap 4, sect 3. part 2 of 2 10 points homework 03 – VENNES, ROSS – Due: Sep 21 2007, 1:00 am 2 b) At what angle with the wall will the particle strike? Correct answer: 50 . 3666 ◦ (tolerance ± 1 %). Explanation: When the particle strikes the wall, the ver- tical component is the side adjacent and the horizontal component is the side opposite the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 7 . 6 m / s 6 . 29474 m / s parenrightbigg = 50 . 3666 ◦ . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (2 . 6 m / s 2 ) (2 . 42105 s) 2 = 7 . 61994 m . Question 3, chap 4, sect 3. part 1 of 1 10 points A quarterback takes the ball from the line of scrimmage, runs backward for 15 . 0 yards, then runs sideways parallel to the line of scrimmage for 13 . 0 yards. At this point, he throws a 40 . 0 yard forward pass straight down the field. What is the magnitude of the football’s resultant displacement? Correct answer: 28 . 178 yards (tolerance ± 1 %). Explanation: Basic Concepts: Δ x tot = Δ x Δ y tot = Δ y 1 + Δ y 2 The displacements are perpendicular, so d tot = radicalBig (Δ x tot ) 2 + (Δ y tot ) 2 Given: Δ y 1 = − 15 . 0 yards Δ x = ± 13 . 0 yards Δ y 2 = 40 . 0 yards Solution: Δ y tot = − 15 yards + 40 yards = 25 yards Δ x tot = ± 13 yards Thus d = radicalBig ( ± 13 yards) 2 + (25 yards) 2 = 28 . 178 yards Question 4, chap 4, sect 4....

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