1. Using Graphs In Physics

2. Find the acceleration of the object for the first 10 seconds of travel Slope = rise/run = (40 m/s)/(10 s) = 4.0 m/s 2

3. Area = ½(20 m/s)(80 m/s) + (10 m/s)(80 m/s) = 800m + 1600m = 2400m Find the distance the object traveled in the first 30 seconds of time.

4. Find the average velocity between 0.0 and 4.0 seconds. Slope of chord = rise/run ≈ 18m/4.0s ≈ 4.5 m/s

5. Find the instantaneous velocity at 4.0 seconds. Slope of tangent line= rise/run ≈ 60m/8.0s ≈ 7.5 m/s

6. Find the instantaneous velocity at 6.0 seconds. Slope of tangent line= rise/run ≈ 84m/6.9s ≈ 12.2 m/s

7. Find the instantaneous velocity at 4.0 seconds. Slope of tangent line= rise/run ≈ 97m/6.0s ≈ 16.2 m/s

8. In the acceleration graph, find the change in velocity between 1.0 and 5.0 seconds. Area = ½bh = ½(5.00s)(12.0 m/s 2 ) = -30.0 m/s (negative - because decelerating)

9. Using Kinematic Equations A hydroplane crosses the finish line traveling at 200.0 meters per second and is decelerated uniformly to rest (0 m/s) in 40.0 seconds. Find the deceleration rate of the watercraft. a = ∆v/t = (v f - v i )/t = (0 m/s-200 m/s)/ 40.0 s = -5.00 m/s 2 Calculate the distance it traveled during the deceleration period. Method 1 x = ½at 2 = ½ (-5.00 m/s 2 )/(40.0s) 2 = 4000m = 4.00 x 10 3 m Method 2 x =v avg * t = ((200.0 m/s + 0 m/s)/2) * (40.0 s) = 4000m = 4.00 x 10 3 m

10. A physics class has a lab assignment to estimate the acceleration due to gravity on the Earth by dropping a ball 2.000 meter to the ground. Each group makes three timing trials to drop the ball one meter and each group averages their timing data. Group A uses a hand stopwatch to time a bowling ball falling 2.000 meter. Group B uses an electronic motion detector connected to a computer to time a falling bowling ball. Group C uses an electronic motion detector connected to a computer to time a falling basketball. The data for each group is shown in the table below.

11. (a) For Group B, use the distance the bowling ball falls and the average falling time to find the average velocity of the bowling ball in miles per hour. Assume 1609 meters = 1.000 mile and 3600 seconds = 1.000 hour. (6 pts)

12. (b) Use the 2.000 m dropping distance and a kinematics motion equation to calculate the theoretical time for a ball to fall to the ground on Earth assuming friction can be neglected. Assume the acceleration due to gravity is 9.80 m/s 2. (8 pts) y = ½gt 2 (solve for time) → t = √((2y)/g) = t = √((2*2.000m)/(9.80 m/s 2 )) = 0.639 s

13. (c) Use the 1.000m dropping distance, the average time to fall, and a kinematics motion equation to calculate the experimental acceleration due to gravity for all three of the groups. (12 pts) y = ½g a t a 2 → (solve for g) g = 2y/t 2 = (2 * 2.000m)/(0.789s) 2 = 6.43 m/s 2 (Group A) = (2 * 2.000m)/( 0.635s) 2 = 9.92 m/s 2 (Group B) = (2 * 2.000m)/( 0.637s) 2 = 9.85 m/s 2 (Group C)

14. (d) Calculate the percentage error of each group’s experimental acceleration due to gravity value assuming the theoretical value of acceleration due to gravity is 9.80 m/sec 2. (3 pts) Group A: ( │ 6.43 m/s 2 - 9.80 m/s 2 │ / 9.80 m/s 2 )* 100 = 34.4% error Group B: ( │ 9.92 m/s 2 - 9.80 m/s 2 │ / 9.80 m/s 2 )* 100 = 1.2% error Group C: ( │ 9.85 m/s 2 - 9.80 m/s 2 │ / 9.80 m/s 2 )* 100 = 0.0% error

15. 29. An arrow is accelerated by a bow from rest to a velocity of 40.0 m/sec. What is the arrow’s acceleration if the arrow was pulled back in the bow 0.500 meters? (8 pts) v 2 = v o 2 + 2ax (solve for a) → a = v 2 / 2x = (40.0 m/s) 2 / (2 * 0.500 m) = 1600 m/s 2

16. Anna kicks a soccer ball into the air and finds the hang-time (total time the ball was in the air) to be 7.20 seconds. What is the highest elevation the ball will achieve above Anna’s head? y = ½gt 2 = ½ (9.80 m/sec 2 )/(3.60s) 2 = ________ m At what vertical velocity will the ball strike a receiver assuming he catches the ball at Anna’s head height? v = gt = (-9.80 m/sec 2 )/(3.60s) = -_________ m/s At what time(s) after the ball is kicked is the ball 15.0 meters above Anna’s head? (c) y = y o + v oy t + ½gt 2 (10.0m) = (20.6 m/s) t + ½(-9.80m/s 2 ) t 2 0 = -20.0 + 20.6 t – 4.90 t 2 t = 1.52 sec on the way up and t = 2.68 sec on the way down.