Sec. 2–2: Acceleration

1. Sec. 2–2: Acceleration Coach Kelsoe Physics Pages 48–59

2. Objectives • Describe motion in terms of changing velocity. • Compare graphical representations of accelerated and nonaccelerated motions. • Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration.

3. Changes in Velocity • In most real-life situations, we rarely have constant velocity. There is usually a speeding up or slowing down period in motion. • This change in velocity is what we call acceleration.

4. Changes In Velocity • Acceleration is the rate at which velocity changes over time. • An object accelerates if its speed,direction, orboth change. • Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

5. Acceleration • The units for acceleration are meters per second squared (m/s2). • We generally use the term “acceleration” to mean “speeding up,” and “deceleration” to mean slowing down. • A negative acceleration indicates deceleration. Remember that accelerations can be positive or negative.

6. Sample Problem B • A shuttle bus slows down with an average acceleration of -1.8 m/s2. How long does it take the bus to slow from 9.0 m/s to a complete stop? • Givens: • vi = 9.0 m/s • vf = 0 m/s • aavg = -1.8 m/s2 • Unknown: • Δt = ?

7. Sample Problem B • We don’t have a formula for time, but we can rearrange our acceleration formula to solve for time: • aavg = Δv/Δt Δt = Δv/aavg • Δt = vf – vi /aavg= 0 m/s – 9.0 m/s/-1.8 m/s2 • Δt = 5.0 s • Look for statements such as “starts at rest” (vi = 0 m/s) and “comes to rest” (vf = 0 m/s) as keys to your given information.

8. I’ll Take the Shuttle • A shuttle bus is at rest and increases its speed to 10 m/s in 5.0 seconds. What is it’s acceleration? • 10 m/s – 0 m/s/5 s = 2.0 m/s2 • The shuttle bus is cruising along at 15 m/s when a dog runs in front of it. It slams on the brakes and comes to a complete stop in 1.5 seconds. What is its acceleration? • 0 m/s – 15 m/s/1.5 s = -10 m/s2 (deceleration)

9. Acceleration • Imagine a high-speed train leaving a station moving to the right so that both displacement and velocity is positive. • The velocity increases in magnitude as the train picks up speed. As the velocity increases, so does the acceleration. • As long as Δv is positive, acceleration will be positive.

10. Acceleration • Imagine the same train slowing down as it approaches the next station. • The velocity is still positive because it is traveling in the same direction. • In this case, the initial velocity is higher than the final velocity, so Δv will be negative and acceleration will be negative. • Can there still be motion with no acceleration?

11. Changes In Velocity • Consider a train moving to the right, so that the displacement and the velocity are positive. • The slope of the velocity-time graph is the average acceleration. • When the velocity in the positive direction is increasing, the acceleration is positive, as at A. • When the velocity is constant, there is no acceleration, as at B. • When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.

12. Velocity and Acceleration

13. Motion with Constant Acceleration • On page 51 of your textbook, there is a strobe photo of a billiard ball falling. There were 10 images taken within one second. • As the ball’s velocity increases, the ball travels a greater distance during each time interval. • The velocity increases by exactly the same amount during each time interval. Thus the acceleration is constant. • The relationships between displacement, velocity, and constant acceleration are expressed by equations that apply to any object moving with constant acceleration.

14. Motion with Constant Acceleration • Displacement depends on acceleration, initial velocity, and time. • As we learned in Section 2–1 , average velocity is equal to displacement divided by time: • average velocity = displacement/time • vavg = (Δx/Δt) • For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity: • average velocity = initial velocity + final velocity/2 • vavg = (vi + vf/2)

15. Motion with Constant Velocity • To find an expression for the displacement in terms of the initial and final velocity, we can set the expressions for average velocity equal to each other: • Δx/Δt = (vi + vf)/2 • Multiplying both sides of the equation by time (Δt) gives us an expression for the displacement as a function of time. This equation can be used to find the displacement of any object moving with constant acceleration. • Δx = ½(vi + vf)Δt • displacement = ½(initial + final velocity)(time interval)

16. Sample Problem • A race car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5 s later. Find the distance that the car travels during braking.

17. Sample Problem Solution • Identify givens and unknowns: • vi = 42 m/s – vf = 0 m/s • Δt = 5.5 s – Δx = ? • Choose correct equation: • The variables we know are initial velocity, final velocity, and time. We don’t know displacement. • Δx = ½(vi + vf)Δt

18. Sample Problem Solution • Plug our values into the equation: • Δx = ½(vi + vf)Δt • Δx = ½(42 m/s + 0 m/s)(5.5 s) • Δx = ½(42 m/s)(5.5 s) • Δx = 120 m • Notice how our units cancel out correctly to give us the units we expect to get.

19. Motion with Constant Acceleration • What happens if the final velocity of an object is not known but we still want to calculate the displacement? • If we know the initial velocity, the acceleration, and the elapsed time, we can find the final velocity. We can then use this value for the final velocity to find the total displacement of the object. • a = Δv/Δt = (vf - vi)/Δt • aΔt = vf – vi aΔx + vi = vf

20. Motion with Constant Acceleration • You can use this equation to find the final velocity of an object after it has accelerated at a constant rate for any time interval. • vf = vi + aΔt • If you want to know the displacement of an object moving with constant acceleration over some certain time interval, you can obtain another useful expression for displacement by substituting the expression for vf into the expression for Δx.

21. Motion with Constant Acceleration • By substituting our vf equation into our Δx equation, we derive a new equation. • Δx = ½(vi + vf)Δt • Δx = ½(vi + vi + aΔt)Δt • Δx = ½[2viΔt + a(Δt)2] • Δx = viΔt + ½a(Δt)2 • This equation is useful not only for finding the displacement of an object moving with constant acceleration but also for finding the displacement required for an object to reach a certain speed or come to a stop.

22. Sample Problem • A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off?

23. Sample Problem Solution • Identify givens and unknowns: • vi = 0 m/s – a = 4.8 m/s2 • Δt = 15 s – vf = ? • Δx = ? • Choose the correct equations: • vf = vi + aΔt • Δx = viΔt + ½a(Δt)2

24. Sample Problem Solution • Plug our values into the equations. • vf = vi + aΔt • vf = 0 m/s + (4.8 m/s2)(15 s) • vf = 72 m/s • Δx = viΔt + ½a(Δt)2 • Δx = (0 m/s)(15 s) + ½(4.8 m/s2)(15 s)2 • Δx = 0 m + 540 m • Δx = 540 m

25. Motion with Constant Acceleration • To this point, all of the equations for motion under constant acceleration have required knowing the time interval. • We can also obtain an expression that relates displacement, velocity, and acceleration without using the time interval. • This method involves rearranging one equation to solve for Δt and substituting that expression in another equation, making it possible to find the final velocity of a uniformly accelerated object without knowing how long it has been accelerating.

26. Motion with Constant Acceleration • We’ll start with one of the equations for displacement: • Δx = ½(vi + vf)Δt  Multiply by 2 • 2Δx = (vi + vf)Δt Divide by (vi + vf) • 2Δx/(vi + vf) = Δt Now use vf = vi + aΔt • vf = vi + a[2Δx/(vi + vf)]  Consolidate the vi • vf - vi = a[2Δx/(vi + vf)]  Multiply by (vi + vf) • (vf - vi)(vf + vi) = 2aΔx Consolidate left side • vf2 - vi2 = 2aΔx Move the vi2 term to the left • vf2 = vi2 + 2aΔx

27. Motion with Constant Acceleration • When using this equation, you must take the square root of the right side of the equation to find the final velocity. Keep in mind that the square root may be either positive or negative. • If you have been consistent in your use of the sign convention, you will be able to determine which value is the right answer by reasoning based on the direction of motion.

28. Sample Problem Final Velocity After Any Displacement A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

29. Sample Problem Solution • Define • Given: • vi = 0 m/s • a = 0.500 m/s2 • Δx = 4.75 m • Unknown: • vf = ? • Diagram: • Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller. The positive direction is to the right.

30. Sample Problem Solution • Plan • Choose an equation or situation: Because the initial velocity, acceleration, and displacement are known, the final velocity can be found using the following equation: vf2 = vi2 + 2aΔx • Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf. vf =±√(vi)2 + 2aΔx

31. Sample Problem Solution • Calculate • Substitute the values into the equation and solve: vf =±√(0 m/s)2 + 2(0.500 m/s2)(4.75 m) vf =±√0 m2/s2 + 4.75 m2/s2 vf =±√4.75 m2/s2 vf = +2.18 m/s • Evaluate • The stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

32. Vocabulary • Acceleration

33. Important Formulas • Average acceleration • a = Δv/Δt • Displacement with Constant Acceleration • Δx = ½(vi + vf)Δt • Δx = viΔt + ½a(Δt)2 • Velocity with Constant Acceleration • vf = vi + aΔt • Final Velocity After Any Acceleration • vf2 = vi2 + 2aΔx