Acceleration and Motion - Understanding Nonuniform Motion and Constant Acceleration

1. Chapter 3 Accelerated Motion

2. Section 3.1 • Essential Questions: • What is acceleration? • How is acceleration different from velocity? • What information can you learn from velocity-time graphs?

3. Section 3-1 Acceleration • Nonuniform Motion Diagrams • You can feel a difference between uniform and nonuniform motion. • Uniform feels smooth. • Nonuniform motion you might feel pushed or pulled.

4. Particle Model Diagram • The change in spacing of the dots and the differences in the lengths of the velocity vectors indicate the changes in velocity. • (Remember, velocity vectors are red!)

5. Displaying Acceleration on a Motion Diagram • For a motion diagram to give a full picture of an object’s movement, it should contain information about the rate at which the object’s velocity is changing. • Acceleration is the rate at which velocity changes. • Acceleration can be either positive or negative. • Acceleration is a vector quantity. • Acceleration vectors are normally violet.

6. Finding Acceleration Vectors • The acceleration is zero in the first example since the velocity vectors are the same size. • The acceleration is positive in the second example since the velocity is increasing. • The acceleration is negative in the last example since the velocity is decreasing.

7. Direction of Acceleration • To find the direction using velocity vectors, we will take the final velocity vector and subtract the initial velocity vector from it.

8. Velocity-Time Graphs • On a velocity-time graph, velocity is plotted on the vertical axis (y) and time is plotted on the horizontal axis (x). • The slope on a velocity-time graph is equal to acceleration. • The area under the curve on a velocity-time graph is equal to displacement.

9. Velocity-Time Graph Constant Velocity = No Acceleration Slowing Down = - Acceleration Speeding Up = + Acceleration Slowing Down in Reverse= + Acceleration Speeding Up in Reverse = - Acceleration

10. Average Acceleration • Average acceleration is a change in velocity over a change in time. • a = v = (v2 - v1) t (t2 - t1) • Units are usually m/s/s or m/s2. • When going in a positive direction: • An increase in velocity is a positive acceleration. • A decrease in velocity is a negative acceleration. • When going in a negative direction: • An increase in velocity is a negative acceleration. • A decrease in velocity is a positive acceleration.

11. Ex #1. The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration? a = v t a = (v2 - v1) (t2 - t1) a = (16 m/s - 2.0 m/s) (4.5 s - 1.0 s) a = (14 m/s) (3.5 s) a = 4.0 m/s2

12. Instantaneous Acceleration • The change in an object’s velocity at an instant of time is called instantaneous acceleration. • You can determine the instantaneous acceleration of an object by drawing a tangent line on the velocity-time graph at the point of time in which you are interested.

13. Acceleration with Constant Speed • Acceleration can occur even when speed is constant. • A change in velocity results in acceleration. • If an object changes direction, but not speed, it accelerates.

14. Section 3.2 Motion with Constant Acceleration • Essential Questions: • What do a position-time graph and a velocity-time graph look like for motion with constant acceleration? • How can you determine the displacement of a moving object from its velocity-time graph? • What are the relationships among position, velocity, acceleration, and time?

15. Position with Constant Acceleration • Constant acceleration (uniform) - acceleration does not change. • The velocity-time graph of constant acceleration produces a straight line. • The position-time graph of constant acceleration produces a curved line.

16. Derive Time! Constant acceleration a = v t a = (vf - vi) t at = vf - vi at + vi = vf vf = vi + at

17. Ex #2. If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of +4.0 m/s2 for 2.5 s, what is its velocity at time t = 2.5 s? vf = vi + at vf = (2.0 m/s) + (4.0 m/s2)(2.5 s) vf = (2.0 m/s) + (10. m/s) vf = 12 m/s

18. We will derive three more equations to use, and only use, when acceleration is constant.

19. Derive Time! Average velocity v = (vf + vi) 2 v = ½ (vf + vi) velocity equals displacement over time d/t = ½ (vf + vi) d = ½ (vf + vi)t

20. Ex #3. What is the displacement of a train as it is accelerated uniformly from +11 m/s to +33 m/s in a 20.0 s interval? d = ½ (vf + vi)t d = ½ (33 m/s + 11 m/s)(20.0 s) d = ½ (44 m/s)(20.0 s) d = 440 m

21. Derive Time! d = ½ (vf + vi)t acceleration equals velocity over time a = v t v = at Substitute (vi + at) in for vf d = ½ ((vi + at) + vi )t d = ½ (2vi + at)t d = vit + ½ at2

22. Ex #4. A car starting from rest accelerates uniformly at +6.1 m/s2 for 7.0 s. How far does the car move? d = vit + ½ at2 d = (0.0 m/s)(7.0 s) + ½ (6.1 m/s2)(7.0 s)2 d = 0 + ½ (6.1 m/s2)(49 s2) d = 149.45 m d = 150 m

23. Derive Time! We know d = ½ (vf + vi)t and vf = vi + at Solve the second equation for t. t = (vf - vi)/a Substitute t d = ½ (vf + vi)[(vf - vi)/a] d = ½ [(vf + vi)(vf - vi)]/a 2ad = [(vf + vi)(vf - vi)] 2ad = [vf2 - vi2] vf2 = vi2 + 2ad

24. Ex #5. An airplane must reach a velocity of 71 m/s for takeoff. If the runway is 1.0 km long, what must the constant acceleration be? vf2 = vi2 + 2ad (71 m/s)2 = (0.0 m/s)2 + 2a(1000 m) 5041 m2/s2 = 0 + (2000 m) a 5041 m2/s2 = (2000 m) a a = (5041 m2/s2) / (2000 m) a = 2.5205 m/s2 a = 2.5 m/s2

25. Equations of Motion for Uniform Acceleration

26. Section 3.3 Free Fall • Essential Question: • What is free-fall acceleration? • How do objects in free fall move?

27. In 1971, astronaut David Scott dropped a hammer and a feather at the same time from the same height above the Moon’s surface. The hammer’s mass was greater, but both objects hit the ground at the same time because the Moon has gravity but no air. (video)

28. Galileo’s Discovery • Free fall is the motion of an object when gravity is the only significant force acting on it. • Galileo was the first to show that all objects fall to Earth with a constant acceleration. • The acceleration of an object due only to the effect of gravity is known as free-fall acceleration. • g = -9.80 m/s2

29. Acceleration Due to Gravity • Acceleration Due to Gravity (g)is another name for free-fall acceleration. • Any object that is dropped will have an initial velocity of zero. • We will assume no air resistance. • An object that is thrown upward will have a velocity of zero at the highest point.

30. Ex #6. The Demon Drop ride at Cedar Point is freely falling at 1.5 s. How far does it fall? What is it velocity at the end of this time? d = vit + ½ at2 We will substitute g in for a d = vit + ½ gt2 d = 0 + ½ (-9.80 m/s2)(1.5 s)2 d = - 11 m vf = vi + gt vf = 0 + (-9.80 m/s2)(1.5 s) vf = -14.7 m/s vf = -15 m/s

31. Variations in Free Fall • As the distance from Earth increases, the effect of free-fall acceleration decreases. • New York City = - 9.81 m/s2 • Denver = - 9.79 m/s2 • Moon’s surface = -1.6 m/s2 • For calculation in this chapter, a value of -9.80 m/s2 will be used for free-fall acceleration.