1. Notes Section 2-2 - VELOCITY and ACCELERATION

2. Speed and Velocity Velocity is not the same as speed.
Speed has magnitude but no direction, thus speed is a scalar quantity.
Since velocities require magnitude and direction for their description and they combine appropriately, velocities are vector quantities.

3. Graphing
If a graph representing velocity is a straight line, the velocity is constant.
A graph representing variable velocity will be a curved line.
The slope of the line of a graph is the ratio of the vertical component of the line to the corresponding horizontal component.

9. Velocity and Acceleration
Velocity measures changes in displacement. vavg = x/ t
Acceleration measures changes in velocity aavg = v / t
Acceleration is the rate of change of velocity in a given time interval.
A complete description of acceleration requires magnitude and direction, and therefore acceleration is a vector quantity.

10. Average Acceleration
Average acceleration = change in velocity / change in time. aavg = v / t or aavg = vf – vi / Δt If velocity, v, is in units of m/s and time, t, is in units of seconds, s, then acceleration, a, is in units of m/s2 aavg = v / t → m/s / s = m/s x 1/s = m/s2 The direction of the acceleration is NOT always the same as the direction of the velocity. The direction of the acceleration depends on the direction of the motion and on whether the velocity is increasing or decreasing

15. Velocity and Accelerationvi a
motion +
Speeding up -
Slowing down
- or +
Constant speed
Speeding up from rest
stationary

16. EX. 1 Find the average acceleration of an amusement park ride that falls from rest to a speed of 28 m/s in 3.0 seconds.
t = 3 s aavg = vf – vi / Δt
vi = 0 m/s
vf = 28 m/s
aavg = ?

17. EX. 2 As a shuttle bus comes to a normal stop, it slows from 9
EX. 2 As a shuttle bus comes to a normal stop, it slows from 9.00 m/s to 0.00 m/s in 5.00 s. Find the average acceleration of the bus. t = 5 s aavg = vf – vi / Δt vi = 9 m/s vf = 0 m/s aavg = ?

18. EX. 3 Find the time required for a motorcycle, with an average acceleration of 15 m/s2, to accelerate from rest to 27 m/s.
t = ? aavg = vf – vi / Δt
vi = 0 m/s
vf = 27 m/s
aavg = 15 m/s2

20. Constant Acceleration
Use these formulas ONLY when the acceleration is constant. For constant acceleration, aavg = a.
VARIABLE QUANTITY SAMPLE UNITS
vf final velocity m/s, km/hr, km/min
vi initial velocity m/s, km/hr, km/min
a acceleration m/s2, km/hr2, km/min2
t time interval s, min, hr
x displacement m, km, cm, dm, µm

21. Equations for constantly
accelerated straight-line motion
Variables in the equation
vf = vi + a t
final velocity, initial velocity, acceleration, time
vf2 = vi2 + 2a x
final velocity, initial velocity, acceleration, displacement
x = vi t + ½ a t2
initial velocity, acceleration, displacement, time
x = ½ (vi + vf) t
final velocity, initial velocity, displacement, time

22. vi = 302 km/hr x 1000 m / 1 km x 1 hr / 3600 sec = vf = 0 m/s
EX. 1 An aircraft has a landing speed of 302 km/hr. The landing area of an aircraft carrier is 195 m long. What is the minimum constant acceleration, in m/s2, for a safe landing?
x = 195 m vf2 = vi2 + 2a x
vi = 302 km/hr x 1000 m / 1 km x 1 hr / 3600 sec =
vf = 0 m/s
a = ? m/s2
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

23. EX. 2 A car can accelerate from 15 m/s to 22 m/s in 2 seconds
EX. 2 A car can accelerate from 15 m/s to 22 m/s in 2 seconds. What is the car's acceleration?
t = 2 s vf = vi + a t
vi = 15 m/s
vf = 22 m/s
a = ? m/s2
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

24. vf = vi + a t. vf2 = vi2 + 2a x. x = vi t + ½ a t2
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

25. EX. 4 A truck driving 14 m/s starts to accelerate at 3 m/s2
EX. 4 A truck driving 14 m/s starts to accelerate at 3 m/s2. After 50 seconds, how far has the truck traveled?
t = 50 s x = vi t + ½ a t2
vi = 14 m/s
a = 3 m/s2
x = ?
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

26. EX. 5 A car on a wet road can achieve an acceleration of only –1
EX. 5 A car on a wet road can achieve an acceleration of only –1.0 m/s2 without sliding. Find the required stopping distance in meters for a speed of 48 km/hr.
vf = 0 m/s vf2 = vi2 + 2a x
vi = 48 km/hr x 1000 m / 1 km x 1 hr / 3600 sec =
a = -1 m/s2
x = ?
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

27. a. What is the velocity of the stroller after it has traveled 4.75 m?
EX. 6 A baby sitter pushing a stroller from rest and accelerates at a rate of m/s2.
a. What is the velocity of the stroller after it has traveled 4.75 m?
b. How long did it take to travel the 4.75 m?
vf = ? (a) vf2 = vi2 + 2a x (b) vf = vi + a t
vi = 0 m/s
a = 0.5 m/s2
x = 4.75 m
t = ?
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t2 x = ½ (vi + vf) t

28. Independent work in textbook
P. 53 Q 1-4
P. 55 Q 1-4
P. 58 Q 3-5
P. 59 Q 1, 3

29. Free Fall and Acceleration
Section 2-3, Free Fall and Acceleration
In the absence of air resistance, all objects dropped near the surface of a planet fall with the same constant acceleration. This motion is referred to as free fall.
Free fall is the motion of an object falling with a constant acceleration  .

30. Free Fall and Acceleration
The acceleration due to the force of gravity or free fall acceleration varies over the earth’s surface. But for most of our problems, we will use the magnitude a = g = 9.81 m/s2.
Since gravity pulls everything toward the earth's surface, the acceleration due to gravity is considered to be pointing down towards the earth's center.

31. Falling Objects faster than
at the same speed as
faster than
In a vacuum, a dense object such as a lead shot falls (slower than, at the same speed as, faster than) an object at lower density, such as a dry tree leaf.
In air, a dense object such as a lead shot falls (slower than, at the same speed as, faster than) an object at lower density, such as a dry tree leaf.
Since acceleration is a vector, its direction can be designated as positive or negative

32. Free Fall Acceleration
If an object is thrown upward and upward displacement is considered a positive direction with a positive velocity, then free fall acceleration will be negative or a = -g = m/s2.
The object is slowing down; velocity and acceleration have opposite signs.

33. Free Fall Acceleration
If an object is dropped from a height and the downward displacement is considered a positive direction, then free fall acceleration will be positive or a = g = 9.81 m/s2.
The object is speeding up or acceleration; velocity and acceleration have the same sign.

35. Acceleration
But what is acceleration at the top of the motion when the object stops? m/s2
vf = vi + a t vf2 = vi2 + 2a x x = vi t + ½ a t x = ½ (vi + vf) t
The above equations were used for objects moving horizontally, where x was used for the displacement.
For objects moving vertically x is replaces with y. So the formulas become:
vf = vi + a t vf2 = vi2 + 2a y y = vi t + ½ a t y = ½ (vi + vf) t

36. Ex. 1 An car accelerates from rest at 3. 00 m/s2 for 10 seconds
Ex. 1 An car accelerates from rest at 3.00 m/s2 for 10 seconds. What is the car's velocity after 10 seconds?
(Is this a free fall problem? No.
What does a equal? a = 3.00 m/s2)

37. Ex. 2 You drop a water balloon out a second story window. 1
Ex. 2 You drop a water balloon out a second story window. 1.1 secs elapses before the balloon hits ground. (a) How high up is the window? (b) How fast is the balloon moving when it hits the ground? y = (a) y = vi t + ½ a t2 (b) vf = vi + a t vi = 0 m/s y = (0)(1.1) + ½ (9.81)(1.1)2 vf = 0+ (9.81)(1.1) t = 1.1 sec y = 5.95 m vf = 10.8 m/s a = 9.81 m/s2 vf =
vf = vi + a t vf2 = vi2 + 2a y y = vi t + ½ a t y = ½ (vi + vf) t

38. Ex. 3 An object is thrown vertically upward with an initial velocity of 12 m/s.
(a) How long will it take the object to reach its maximum height? (b) What is the maximum height that the object achieved? (c) What will be the object's velocity when it returns to its starting point?
y = (b) y = vi t + ½ a t2 (a) vf = vi + a t
vi = 12 m/s y = (12)(1.22) + ½ (-9.81)(1.22)2 0 = (-9.81)( t)
t = y = 7.34 m t = 1.22 sec
a = m/s2
vf = 0 m/s (c) vf = -12 m/s

39. Ex. 4 An object is thrown upward with an initial velocity of 5. 0 m/s
Ex. 4 An object is thrown upward with an initial velocity of 5.0 m/s. How long will it take the object to reach its starting point? t = vf = vi + a t vi = 5 m/s -5 = 5 + (-9.81)( t) vf = -5 m/s t = 1.02 sec a = m/s2