1 topics distance, location, speed speed and direction directional quantities acceleration free fall...
TRANSCRIPT
1
Topics
• Distance, Location, Speed
• Speed and Direction
• Directional quantities
• Acceleration
• Free Fall
• Graphs of Motion
• Derivatives and Integrals
2
Average Speed
• distance: total path length
• speed: rate of travel (e.g. 50 mph)
• Average Speed: distance/time (e.g. 100m in 3.0s)
[m/s] timetravel
distances
3
Displacement: Change in Position
if xxx SI Unit: meters (m)
4
Velocity (m/s)
0 : : velocityaverage
tt
xvavg
0 : : velocityousinstantane
tt
xv
5
Velocity Examples
• average velocity: 60mph toward Dallas
• instantaneous velocity: 11:47am: Northbound, 83mph
6
Example: Average Velocity
to = 0.0s, xo = 5.0m, vo = +2.0m/s
t = 1.2s, x = 3.08m, v = -5.2m/s
smss
mm
t
xvavg /6.1
0.02.1
00.508.3
Note that velocities always have directional information. Here the “-” sign means –x direction.
7
Scalars & Vectors
• Scalar: size only
• e.g. speed, distance, time
• Vector: magnitude and direction
• e.g. displacement, velocity, acceleration
8
A honeybee travels 2 km round trip before returning. Is the displacement for the trip the same as the distance traveled?
1 2
79%
21%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
41 42 43 44 45
1. Yes
2. No
9
Acceleration (m/s/s)
0 : :onaccelerati average
tt
vaavg
0 : :onaccelerati ousinstantane
tt
va
10
Example: Car goes from 10m/s to 15m/s in a time of 2.0 seconds. Calculate the average acceleration.
m/s/s 5.20.0s-2.0
10m/s-15
t
vaavg
11
Previous Example:
to = 0.0s, xo = 5.0m, vo = +2.0m/s
t = 1.2s, x = 3.08m, v = -5.2m/s
m/s/s 0.60.0s-1.2
2.0m/s-5.2-
t
vaavg
12
Motion Diagrams
• velocity arrow and position• zero velocity is a “dot”• acceleration & net-force directions: parallel to v• Example: slowing, reversing direction
13
Kinematic Equations of Constant Acceleration
atvv o :velocity
tvvx o )( : velocityaverage 21
221 :ntdisplaceme attvx o
xavv o 2 :squared-v 22
14
Displacement and x vs. t Graph
15
x vs. t Graph
• slope is velocity
16
v vs. t Graph• slope is acceleration
atvv o
17
Human Acceleration
mat 20221
In the 1988 Olympics, Carl Lewis reached the 20m mark in 2.96s. Calculate average acceleration.
20)96.2( 221 a
ssms
ma //56.4
)96.2(
2022
18
Cheetah Acceleration
A cheetah can accelerate from 0 to 20m/s in 2.0s. What is the average acceleration?
ssms
sm
t
vva o //10
0.2
/)020(
19
Ex: V2 EquationApproximate Stopping Accelerations in m/s/s:
Dry Road: ~ 9 (anti-lock) ~ 7 (skidding)
Wet Road: ~ 4 (anti-lock) ~ 2 (skidding)
At 60mph = 27m/s, what is the stopping distance of a skid on a wet road?
feet) 006(about 182
)2(2270
222
22
mx
x
xavv o
20
Free-Fall
• only gravity acts
• air-friction is negligible
• a = 9.8m/s/s downward
21
Calculus of Linear Motion
• derivatives and integrals
• Examples:
• dx/dt = v dv/dt = a
• d/dt(3 + 4t + 5t2) = 4 + 10t
• v = integral of acceleration
22
Velocity
a(t)dtvv o
22545 ttdtt
Example:
23
Summary:• speed: rate of travel• average speed: distance/time.• displacement: change in position• velocity: rate position changes• acceleration: rate velocity changes• kinematic equation set• free fall: constant acceleration.• graphs and slopes• derivatives and integrals of polynomials
24
25
Example: A solid metal ball is projected directly upward with velocity +5.0m/s. How high does it go? How long does it take to return to same height?
mh
gh
gh
yavv o
28.1
6.19/25)2/(25
250
222
22
sgt
gt
t
gttgtt
attvy o
02.18.9/10/10
0)5(
0
)5(50
21
212
21
221
26
Case Study: 100 meter track-race
1. a = const., 0-60 m 2. top speed of 16 m/s at 60 m. 3. a = 0, 60-100 m
velocity vs time
0.002.004.006.008.00
10.0012.0014.0016.0018.00
0.00 2.00 4.00 6.00 8.00 10.00 12.00
t(s)
velo
city
(m/s
)
27
st
t
t
at
attvx
t
o
5.78/60
860
60
060216
21
221
221
2/13.25.7/16
16
016
sma
ta
at
atvv o
a) Acceleration and Time
100m Race
28
st
t
attvx o
5.216/40
01640
221
b) Time and Distance: Last 40meters of race at constant speed of 16m/s.
Race Time = tI + tII = 7.5s + 2.5s = 10.0s
100m Race
29
v = vo + at.16 = 0 + a(7.5)a = 16/7.5 = 2.13 m/s2.
c) We can also use time found in part (a) in velocity equation to get the acceleration of the runner in 1st part of the race.
x = vavgt = {(vo + v)/2}t = {(0 + 16)/2)}(7.5) = (8)(7.5) = 60m.
d) Distance using vavg
30
Position vs time
0.00
20.00
40.00
60.00
80.00
100.00
120.00
0.00 2.00 4.00 6.00 8.00 10.00 12.00
t(s)
po
sit
ion
(m)
31
Example: An object has velocity of +2.0m/s at x = 5.0m and at t = 0.0s. At t = 1.2s it has velocity of -5.2m/s and position x = 3.08m.
Average Acceleration:
ssmss
smsm
t
vaavg //0.6
0.02.1
/0.2/2.5
smsssmsmatvv o /2.5)2.1)(//0.6(/0.2
Using v(t) equation:
Consistent answer:
How long did it take the object to reach v = 0?
sssm
sm
a
vt
atv
o
o
33.0//0.6
/0.200
0
32
33
A train moves along a straight track. The graph shows the position as a function of time for this train. Note that the speed at an instant is the slope of the line at any point on the line. The graph shows that the train:
1 2 3 4
11%
22%
39%
28%
1. speeds up all the time.
2. slows down all the time.
3. speeds up part of the time and slowsdown part of the time.
4. moves at a constant velocity.
time
position
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
41 42 43 44 45
34
A car travels West at 20m/s. It begins to slow. Use the convention that East is +x. The acceleration of the car is considered positive since if it slowed to 19m/s in 1.0s, then
ssms
sm
t
vva o //1
1
/)20(19
Motion Diagram:
v
v(t)
a
+-
Motion Diagram Example
35
Example: A car starts from rest and travels West with uniformly increasing speed. Use the convention that East is +x. Is the acceleration + or -? Is the total force acting on the car + or -? Draw a motion diagram.
Assume it goes from 0 to -10m/s in 10s.
ssms
sm
t
vva o //1
10
/)0(10
Net-force parallel to acceleration, i.e. force is – direction.
motion diagram
Net Force, Acceleration, & Motion Diagrams
36
A car can accelerate at 6m/s/s. The time to go from 40mph to 60mph is:
smmi
m
s
h
h
mi/87.17
1
1609
3600
140 sm
mi
m
s
h
h
mi/81.26
1
1609
3600
160
atvv o
sssm
sm
a
vvt o 49.1
//6
/87.1781.26
Example using Acceleration
37
VehicleAverage Stopping Distance at 55 mph (includes reaction time)
Passenger car 190 ft.
Tractor-trailer (loaded) with cool brakes
256 ft.
Tractor-trailer (loaded) with hot brakes
430 ft.
Tractor-trailer (empty) 249 ft.
Tractor only (bobtail) 243 ft.
38
VehicleStopping Distancefrom 60 mi/hr
Accel.
feet meters ft/s2 m/s2
BMW M3 120 37 32.3 9.8
Dodge Colt GL 167 51 23.2 7.1
39
40
Time to Stop
BMW
st
t
atvv o
75.2
8.9270
st
t
atvv o
80.3
1.7270
Colt
41
y and v graphs for tossed object in “free-fall”
42
Determine how realistic 6m/s/s is for a car by computing the 0 to 60mph time:
Good time, but can be done.
sssm
sm
a
vvt o 46.4
//6
/081.26
Realistic Car?