Physics Homework #2: Acceleration

Physics 110 Homework #2 (25 points) (In the following questions assume that air resistance is negligible and you can use g = 10m/s² in calculations) Which of the following are examples of constant acceleration? b. A car exiting a highway (graph b) c. A car entering a highway (graph c) f. b and c only The acceleration due to gravity is approximately -10m/s². This means that: b. The velocity of falling objects changes by -10m/s every second In a laboratory on Earth, all the air is pumped from a large tube. A feather and a steel ball are simultaneously released from rest inside the tube. Which will hit the ground first?   c. They will reach the ground at the same time A ball is thrown straight up in the air. At the highest point the velocity will be: a. 0m/s A ball is thrown straight up in the air. At the highest point the acceleration will be: c. -9.8m/s² Physics 110 Homework #2 (Continued) A ball is thrown straight up in the air. At the highest point the acceleration will be: c. -9.8m/s² Which of the following is true about the horizontal and vertical velocities of a projectile? b. The vertical velocity changes by approximately 10m/s every second, the horizontal velocity doesn't change Two balls are released from the same height. One is dropped and the other is projected horizontally. Which will have the larger total velocity just before they hit the ground?   c. The balls will have the same total velocities A car moving at 20m/s accelerates at 3m/s² for 4 seconds. What will be the velocity? c. 32m/s What will be the distance traveled by the car in the previous problem? d. 104m Calculations: Question 8: Final velocity = initial velocity + (acceleration x time) = 20m/s + (3m/s² x 4s) = 32m/s Question 9: Distance = initial velocity x time + 0.5 x acceleration x time² = 20m/s x 4s + 0.5 x 3m/s² x (4s)² = 80m + 24m = 104m Physics 110 Homework #2 (Continued) A ball is dropped from a window, what will be the approximate velocity 4 seconds later? d. -40m/s What approximate distance will the ball in the previous problem have fallen in 4 seconds? c. -80m A ball is thrown straight up in the air with a velocity of 30m/s. What will be the approximate velocity 5 seconds later? b. -20m/s What will be the approximate height of the ball in the previous problem? b. 25m A ball is thrown straight up in the air with a velocity of 30m/s. Approximately how long will it take to reach the highest point?   c. 3s Calculations: Question 10: Final velocity = initial velocity + (acceleration x time) = 0m/s + (-10m/s² x 4s) = -40m/s Question 11: Distance = initial velocity x time + 0.5 x acceleration x time² = 0m/s x 4s + 0.5 x (-10m/s²) x (4s)² = -80m Question 12: Final velocity = initial velocity + (acceleration x time) = 30m/s + (-10m/s² x 5s) = -20m/s Question 13: Height = initial velocity x time + 0.5 x acceleration x time² = 30m/s x 5s + 0.5 x (-10m/s²) x (5s)² = 150m - 125m = 25m Question 14: Final velocity = 0m/s (at the highest point) 0m/s = 30m/s + (-10m/s²) x time 10m/s² x time = 30m/s time = 30m/s / 10m/s² = 3s Physics 110 Homework #2 (Continued) A ball is thrown straight up in the air with a velocity of 30m/s. Approximately how long will it take to reach the highest point? c. 3s A projectile is thrown with an initial horizontal velocity of 30m/s and vertical velocity of 30m/s. After 2 seconds, what will be the approximate horizontal and vertical velocities? b. Horizontal velocity: 30m/s, vertical velocity: 10m/s A projectile is thrown with an initial horizontal velocity of 30m/s and vertical velocity of 30m/s. After 2 seconds, what will be the approximate horizontal and vertical distance traveled? b. Horizontal distance: 60m, vertical distance: 40m Calculations: Question 14: Final velocity = 0m/s (at the highest point) 0m/s = 30m/s + (-10m/s²) x time 10m/s² x time = 30m/s time = 30m/s / 10m/s² = 3s Question 15: Horizontal velocity remains constant: 30m/s Vertical velocity = initial vertical velocity + (acceleration x time) = 30m/s + (-10m/s² x 2s) = 10m/s Question 16: Horizontal distance = horizontal velocity x time = 30m/s x 2s = 60m Vertical distance = initial vertical velocity x time + 0.5 x acceleration x time² = 30m/s x 2s + 0.5 x (-10m/s²) x (2s)² = 60m - 20m = 40m