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Motion in One Dimension
Chapter 2
Mechanics
• Mechanics is the study of motion.• Kinematics is the science of describing the motion of
objects with words, diagrams, numbers, graphs, and equations.
• Terms: Distance Position Displacement Speed Velocity
Average speed Average VelocityAccelerationFreefall Acceleration due to gravity
2.1 Displacement and Velocity
• One-dimensional motion takes place in one direction (straight line motion).
• Frame of reference: – a system for specifying the precise location of objects in space and
time– used to measure changes in position– remains fixed for the motion problem– an origin (starting point) from which motion and time are measured
For the gecko the meter stick is the frame of reference.
2.1 Displacement and Velocity
• Displacement – the change in position of an object and the direction of change
– The length and direction of the straight line drawn from an object’s initial position (xi) to the object’s final position (xf ).
– x = xf –xi or x = xf –x0 The SI unit is the meter, m.
– Motion is relative displacement.
– Speed is the time rate of motion. Speed has no direction.
2.1 Displacement and Velocity• Displacement
– the change in position of an object and the direction of change
– x = xf –xi or x = xf –x0
What is the skater’s displacement?
What is the skater’s distance traveled?
x = + 140 m
d = 420 m
2.1 Displacement and Velocity
• Displacement has direction and sign.
Motion to the right or east is positive (+).
Motion to the left or west is negative (-).
Motion to the north or up is positive (+).
Motion to the south or down is negative (-).
Chapter 2Positive and Negative Displacements
2nd
3rd 4th
2.1 Displacement and Velocity
• Velocity – Average velocity is the total displacement divided by the time
interval during which the displacement occurred– SI Unit: m/s
– can be positive or negative depending on the sign of the displacement
– Average velocity is equal to the constant velocity needed to cover a certain displacement in a given time interval; it is not necessarily the average of the initial and final velocities. (It is equal to the average of the initial and final velocities when acceleration is constant.)
vavg = = = x
txf - xi
t
xf – x0
t
2.1 Displacement and Velocity
• A car travels from A to B which is 100 km. If the first half of the distance is driven at 50 km/h and the second half is driven at 100 km/h, what is the average velocity?
Know:
vavg =x
t
x = 100 kmWhat is t?
t =x
vavg
t0-50km = = = 1.0 hx
vavg
50km
50 km/h
t 50km – 100 km = = = 0.5 hx
vavg
50km
100 km/h
vavg = = = 67 km/hx
t
100 km
1.5 h
2.1 Velocity and Speed
• Velocity describes motion with both a direction and a numerical value (a magnitude).
• Speed has no direction, only magnitude.
• Average speed is equal to the total distance traveled divided by the time interval.
distance traveledaverage speed =
time of travel
2.1 Displacement and Velocity
• p. 44 #1
vavg = + 0.98 m/s
t = 34 min
x = ?
t = 34 min x 60 s/min = 2040 s
vavg =x
t
x = vavgx t
x = 0.98 m/s x 2040 s
x = 1999.2 m
x = 2.0 x 103 m East
2.1 Displacement and Velocity
• p. 44 #2, #4, and #6
2.1 Interpreting Motion Graphically• For any position-time graph, we can determine the average velocity
by drawing a straight line between any two points on the graph.• If the velocity is constant, the graph of position versus time is a
straight line. The slope indicates the velocity.
Time (s)
B
A
C
Pos
ition
(m
)
Object 1: positive slope = positive velocity
Object 2: zero slope= zero velocity Object 3: negative slope = negative
velocity
2.1 Interpreting Motion Graphically
(m)
(s)
2.1 Displacement and Velocity
• Position-Time Graphs (Distance-Time Graphs)– Describe the motion for each graph (1, 2, and 3)
1 2
3
2.1 Displacement and Velocity• Position – Time graphs illustrate velocity.
A
B C
D
E
What is the total displacement of the object? the total distance traveled?
Calculate vavg for AB, BC, CD, DC, AC, AD, and AE.
vavg, AB = +6 m/s
vavg, BC = 0 m/s
vavg, AD = -1 m/s
vavg, CD = -4 m/s
vavg, DE = 2.7 m/s
vavg, AC = 4 m/s
vavg, AE = 0 m/s
2.1 Displacement and Velocity
• Instantaneous Velocity – is the velocity of an object at some instant or at a specific point in the
object’s path.– is equal to the slope of the tangent line at a specific point (instant in
time).
2.2 Acceleration = Change in Velocity
• Acceleration is the change in velocity. • An object accelerates if its speed, direction, or both change.
• Acceleration is the rate at which velocity changes over time.
f iavg
f i
v vva
t t t
change in velocityaverage acceleration =
time required for change
• Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.
2.2 Acceleration
SI units: m/s2
Acceleration has direction and magnitude.
m/ss( )
2.2 Acceleration • Consider a train moving to the right, so that the displacement
and the velocity are positive.
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 ((-) slope), the acceleration is negative, as at C.
• The slope of the velocity-time graph is the average acceleration.
2.2 Positive or Negative Acceleration
Velocity Acceleration Motion
+ + speeding up in + direction
- - speeding up in - direction
+ - slowing down in + direction
- + slowing down in - direction
- or + 0 constant velocity
0 - or + speeding up from rest
0 0 at rest
2.2 Constant Acceleration
2.2 Motion Graphs
Time (s)
Acc
ele
ratio
n(m
/s2)
Displacement -Time Graphslope = m/s velocityarea = m x s meaningless
Velocity -Time Graphslope = m/s/s accelerationarea = m/s x s = m displacement
Acceleration-Time Graphslope = m/s2/s meaninglessarea = m/s2 x s = m/s velocity
2.2 Constant Acceleration
Displacement depends on acceleration, initial velocity, and time.
The time interval between each image is the same. Strobe photograph with 0.1s time intervals.
Why does the ball get farther and farther apart in the same amount of time?(The distance between the images increases while the time interval remains constant.)
The ball is accelerating with constant acceleration due to gravity.
2.2 Constant Acceleration Velocity-Time Graph
The relationships between displacement (x) , velocity (v), and constant acceleration (a) are expressed by equations that apply to any object moving with constant acceleration.
vavg =x
t
vi + vf
2vavg =
vi
vf
*
*
*vavg
Constant acceleration
• At constant acceleration
2.2 Displacement with Constant Acceleration
vavg = and vavg =x
t
vi + vf
2
Using these two equations derive an expression for displacement.
x
t= ½(vi + vf)
x = ½(vi + vf)t
= ½(vi + vf)
2.2 Final Velocity and Acceleration
• From the definition of acceleration calculate final velocity, vf, from initial velocity, acceleration, and time.
a = vf – vi
t
Solve for vf .
at = vf – vi
at + vi = vf
vvf f == aat t + v+ vi i
2.2 Displacement with Constant Acceleration
• Displacement with Constant Acceleration when vf is not known.
x = ½(vi + vf)t vf = at + vi
x = ½(vi + at + vi )t
x = ½(2vit + at2)
x = vx = viit + ½at + ½att22
This equation is useful for finding the displacement required for an object to reach a certain speed. Use vvf f == aat t + v+ vi i to determine t and then this formula for x.
2.2 Final Velocity w/o t at Constant Acceleration
• If t is unknown, the final velocity depends on initial velocity, acceleration, and displacement vf = at + vi
Solvex = ½(vi + vf)t for t and substitute it into the equation
above.
x = ½(vi + vf)t
2x = (vi + vf)t
2x (vi + vf)
= t
vf = at + vi
vf = a + vi 2x
(vi + vf)
(vf - vi ) (vi + vf ) =2ax
vf 2 = vi
2 + 2ax
vf - vi = a 2x (vi + vf)
2.2 Formulas when Acceleration is Constant
x = ½(vi + vf)t x = ½(vf)t
vf = vi + at vf = at
x = vit + ½at2 x = ½at2
vf 2 = vi
2 + 2ax vf 2 = 2ax
Object initially at rest, vi = 0
Yea let’s PRACTICE!!
2.3 Falling Objects
2.3 Falling Objects• Free fall is the motion of any object under the
influence of gravity only, i.e. no air resistance (friction) of any kind.
• Any falling object's motion is at least approximately free fall as long as:1. The object is relatively heavy compared to its size.
2. The object it falls for a relatively short time.
3. The object it is moving relatively slowly.
(This means air resistance depends on speed, size, and shape.)
• An object doesn't have to be falling to be in free fall. If you throw a ball upward its motion is still considered to be free fall, since it is moving under the influence of gravity.
2.3 Falling Objects
• Free fall is motion at constant acceleration.
• ag = acceleration due to gravity
• g = acceleration due to Earth’s gravity
• g = 9.81m/s2 = 32 ft/s2 at Earth’s surface (about 22 mi/hr/s)
• ag = -g = - 9.81m/s2 = - 32 ft/s2 (choose up (+) down is (-)
• For objects thrown up into the air, acceleration is constant during upward and downward motion. Acceleration is -9.81 m/s2 when the object is going up and when the object is coming down.
2.3 Falling ObjectsAcceleration of gravity, "g" in the Solar SystemAt the surface of g (m/s2) is Mass (kg)* Radius (m)*
Sun1 275 1.99 x 1030 6.95 x 108
Mercury 3.7 3.30 x 1023 2.44 x 106
Venus 8.9 4.87 x 1924 6.05 x 106
Earth 9.8 5.97 x 1024 6.38 x 106
Moon (of Earth) 1.6 7.35 x 1022 1.73 x 106
Mars 3.7 6.42 x 1023 3.40 x 106
Jupiter1 25 1.90 x 1027 7.15 x 107
Ganymede (moon of Jupiter)
1.4 1.48 x 1023 2.63 x 106
Europa (moon of Jupiter) 1.3 4.80 x 1022 1.57 x 106
Saturn1 10.4 5.68 x 1026 6.03 x 107
Uranus1 8.9 8.68 x 1025 2.56 x 107
Neptune1 11 1.02 x 1026 2.48 x 107
Pluto 0.7 1.27 x 1022 1.14 x 106
1Assuming that these objects actually had a surface... *Mass and radius data taken from The Nine Planets
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Kinematics/ff_velocity_acc.htm
2.3 Falling Objects
As the object moves up it decelerates, so its velocity decreases. At the peak of the object’s path it stops and turns around. The object’s velocity then begins to increase in the negative direction.
• Velocity – Time Graph• What does it tell us? What does the slope indicate?• What type of motion is represented?
2.3 Falling Objects
Falling Object
Jason hits a volleyball so that it moves with an initial
velocity of 6.0 m/s straight upward. If the volleyball
starts from 2.0 m above the floor, how long will it be
in the air before it strikes the floor?
Sample Problem
2.3 Falling Objects• Terminal Velocity
– constant speed an object reaches when falling through a resisting medium (air)
– depends on the the object’s mass, shape, and size
2.3 Falling Objects
p.64, #1x = 239 m
ag = -3.7 m/s2239 m
vi = 0
vf = ?
t = ?
a. vf
2 = vi 2 + 2ax
vf 2 = 2ax
vf 2 = 2(-3.7m/s2)(-239m) = 1768.6 m2/s2
vf = - 42 m/s
b. a =vf - vi
tt = =
vf
a
-42 m/s
-3.7m/s2= 11s
2.3 Falling Objects
• p.64, #3
Chpt. 2 Reading Guide Keyp. 47 #3
20.0 s22 s
a. vavg = 50.0 m/20.0 s = 2.50 m/s
vavg =x
t
b. vavg = 50.0 m/22.0 s = 2.27 m/s
c. vavg = 0.0 m/s
Chpt. 2 Reading Guide Key
• p. 47 #4
v1 = 0.90 m/s
v2 = 1.90 m/s
780 m vavg =x
t
a. t1 = = 780 m / 0.90m/s = 866.67 s
t2 = = 780 m / 1.90m/s = 410.53 s
866.67 s - 410.53 s = 456 s = 460 s
x
v1
x
v2
Chpt. 2 Reading Guide Key
• p. 47 #4
v1 = 0.90 m/s
v2 = 1.90 m/s
? m vavg =x
t
b. 866.67 s - 410.53 s = 456 s = 460 s
x
v1
x
v2
- = 330 s
t1 – t2 = 5.50 min = 330 s
v1
v2
- = 330 sx
0.90
1.90
- = 330 sx
x = 564.3 m = 560 m
Chpt. 2 Reading Guide Key
p. 47 #6
Bear B It went farther in the same amount of time.
Bear A At 8 min, the slope for Bear A is steeper.
No
No
