Chapter 1 and 2 Study Guide for Physics Unit 1 Test

1. Chapter 1 and 2 Study Guide for Physics Unit 1 Test Mrs. Ashley

2. Vectors and Scalars • Has magnitude but no direction • Speed • Mass • Temperature • Magnitude of a vector • Has magnitude and direction • Velocity • acceleration Scalar Vector

3. Problem #1 • A baseball player runs 27.4 meters from the batter's box to first base, overruns first base by 3.0 meters, and then returns to first base. Find the total distance traveled by the player. What is the magnitude of the player's total displacement from the batter's box?

4. Solving Problem #1 • Distance (total) = d1 + d2 + d3 • Total displacement • Total Distance is: 27.4 meters + 3.0 m + 3.0 m = 33.4 m • Total displacement =(taking into account direction) d1 + d2- d3 • 27.4 m + 3.0 m -3.0 m • Total displacement is 27.4 m from the batters box.

5. Simple Formulas • Speed = distance/time • Average Speed = total distance/time interval • Velocity = Δ displacement/ Δ time • Acceleration = change in velocity/time interval (rate of change in direction or change in speed or both is acceleration) • A curved path is always a change in direction, so it is accelerating even if speed doesn’t change

6. Problem #2v = d/t or d =vt or t = d/v • How long will it take an object to move 100 meters if the object is traveling with an average speed of 0.5 meter per second?

7. Solution to Problem #2 • t= d/v • t = 100m/.5 m/s = 200 s

8. Instantaneous Velocity and Acceleration and Average Acceleration • Velocity = Δ displacement/ Δ time • Average acceleration= velocityf- velocity i • time f - time i

9. Problem #3 • A car is traveling along a straight highway with a velocity of 30 m/s when the driver hits the brakes when a car pulls out in front of it. If it takes the car 5 seconds to slow down to a velocity of 3.0 m/s, what is the car’s average acceleration?

10. Solution to Problem #3 • average acceleration = change in velocity • elapsed time • a = 3.0 m/s – 30 m/s • 5 s  • a = 27 m/s • 5s • a = -5.4 m/s2

11. Derived Formulas for Velocity, Acceleration, Speed and Distance • Constant velocity means that an object is moving the same distance in the same time in the same direction • Instantaneous velocity =acceleration X time v=at • Average velocity = beginning velocity + ending velocity divided by 2, if acceleration is constant • Distance traveled = ½ at2 (if velocity is zero to start)

12. Constant Acceleration • When we want to find final velocity when we know initial velocity, acceleration, initial distance and final distance: • v2f = v2i + 2a( df- di)

13. Constant Acceleration • If we what to find the velocity of an object after a certain amount of time and know the objects constant acceleration we use this formula: • vf = vi + at

14. Constant Acceleration • If we want to find the final position and know the original position, the velocity, time and acceleration we use the following formula: • df= di+ vit+ 1/2 at2

15. Constant Acceleration • If we want to find final distance and know the original position, original velocity, final velocity, and acceleration then we use: • df = di + v2f - v2i • 2a • This formula can be used for braking distance, for example

16. Constant Acceleration • Formula for stopping distance of a car: (tr is reaction time and notice that a is negative) • df = vitr- vi2/ 2a

17. Problem #4 • A car slows down from 40 m/s to a stop in a distance of 100 m. What was its acceleration, assumed constant?

18. Problem #4 solution • v2f = v2i + 2a( df- di) • a = v2f - v2i • 2 ( df- di) • a= 0- (40 m/s )2 • 2(100m) • a = -8 m/s2

19. Problem #5 • If a car slows down to rest uniformly and stops in 63.0 m and its original speed was 21 m/s How long did it take to come to a complete stop?

20. Problem #5 Solution • df= di + average velocity times t • df= di + vi + vft 2 • t = 2 (df- di) • vi + vf • t = 5.5 s or 6 s

21. Problem #6 • Determine the stopping distance for a car with an initial velocity of 80 km/h and a human reaction time of 1.0 s for an acceleration of -6.0m/s2

22. Problem #6 Solution • The origin is the location of the car at the beginning of the reaction time: You have different units, so need to change everything to m/s: • (80km/h) (1h/3600s) ( 1000m/km) = 22.2m/s • d = vt (22.2m/s) (1.0 s) = 22.2 m • df = di + v2f - v2i • 2a

23. Problem #6 Solution • df = 22.2m + 0- 22.2m/s2x -22.2m/s2 • 2(6.0 m/s2) • df = 22.2m+492.84/12 m • Df = 22.2m + 41.07 = 63.27 m= 63 m

24. Free Fall Formulas • Regardless of mass, acceleration is at the same rate 9.8 m/s2 This means that acceleration is 9.8 meters per second in each second of fall. • Free fall acceleration is due to gravity so “g” is used in the formulas • Velocity = gravity/time • Distance = vot+ ½ gt2

25. Free Fall Formulas • If you know the original velocity and the time, you can figure the velocity at the time elapsed by this formula: • vf = vo+ gt

26. Throwing something up • An object slows at 9.8 meters per second squared • Instantaneous speed is same on the way up as the way down at the same position. • Up is usually expressed as a positive and down is expressed as a negative in terms of acceleration due to gravity

27. Problem #7 • If a ballplayer catches a ball 3 seconds after throwing it vertically upward, with what speed did he throw it and what height did it reach?

28. Problem #7 Solution • Upward is the positive direction, then d initial = 0. The acceleration is -9.81 m/s2 The displacement is 0, assuming the ball was caught at the exact location from where it was thrown. Use y as the height. • Yf= Yo+ vot + 1/2 at2

29. Problem #7 Solution • Yf= Yo+ vot + 1/2 at2 • vo =-1/2 at • vo=15 m/s

30. Problem #7 Solution • df = di + v2f - v2o • 2a • df = 0 + 0- (14.7 m/s)2 • 2(-9.81m/s2 • df = 1 m