Download Chapter 7: Velocity, Acceleration and Force

Chapter 7: Velocity, Acceleration and Force
When a stock car is racing at the Kentucky Motor Speedway it
will have periods of time racing at maximum speed; other
periods of time it will be racing under caution flags and in the
pits. The race may take about 2 hours and 30 minutes. The
average speed of the race car can be determined by taking the
distance covered and dividing it by the time to cover the
distance.
Figure 7.1 Frank
Kimmel
Objectives:
•
•
•
•
To explain and solve velocity and speed problems
To explain and solve accelerated motion problems
To graphically understand complex motions
To understand Newton’s Laws of Motion
Speed and velocity
s=
d
t
(7.1)
This is an average speed. At any time during the race the car might be stopped in the pits
or on the straightaway traveling at nearly 200 miles per hour. The instantaneous speed of
the race car can be determined at any time during the race, by using very short distance
and measure the time to cover this distance.
Distance
(m)
533.333
1066.666
1599.999
2133.332
2666.665
3249.998
3933.331
4716.664
5549.997
6383.33
7216.663
8049.996
8883.329
9616.662
10149.995
10483.328
Time (s)
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
Distance
(m)
10616.661
10616.661
10616.661
10616.661
10666.661
10816.661
11066.661
11416.661
11866.661
12416.661
13066.661
13766.661
14466.661
15166.661
15866.661
16566.661
Time (s)
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
The graph is a plot of the distance covered plotted against the time of travel.
Race Car
18000
16000
14000
distance (m)
12000
10000
8000
6000
4000
2000
0
0
50
100
150
200
250
300
350
time (s)
Figure 7.2 Distance vs Time Graph
The slope of a linear portion of the graph is the speed of the car. You can see that other
portions of the graph are composed of either horizontal or curved lines. The curved
portion of the graph represents when the car is accelerating. The acceleration may be a
speeding up or slowing down. A slowing down of the car is a negative acceleration also
known as deceleration. The horizontal portion of the graph is when no distance is
covered during a period of time, thus the car is stopped.
A car which is traveling at a constant velocity has an acceleration of zero. A car which
has an increasing or decreasing velocity is accelerating. Acceleration will be kept
constant for cases examined in this class.
According to Newton’s Second Law of Motion an unbalanced force is required to cause
an object to accelerate assuming that the mass of the object is constant which is true as
long as we don’t worry about the fuel being burned by the car:
F = ma
(7.2)
During the race when the car is traveling at a constant speed the car is creating a force.
This force is equal and opposite to the friction of the tires on the race track. When the car
is accelerating (speeding up) the force propelling the car forward is greater than the
frictional force of the tires on the track. When the car is decelerating the frictional force
is greater than the force the car is producing.
Velocity is defined as speed with a direction. It is a vector quantity. A vector is a
mathematical device that has both magnitude and direction. Acceleration is also a vector
quantity. According to Newton’s First Law of Motion an object in motion remains in
motion in a straight line unless acted upon by an outside force. If the car is traveling on
the straightway it may have a constant velocity, but when the car is going around a curve
the car cannot have a constant velocity since it is not traveling in a straight line. The
outside force is the frictional force between the tires and the road and any banking of the
track. Without the frictional forces and banking of the track the car would continue in
straight-line motion like a car on ice. If the car has force(s) acting on it then according to
Newton’s Second Law of Motion there must be an acceleration present, equation (7.2).
One of the types of accidents that occur in racing is when the car goes straight into the
wall instead of turning. This occurs when the tires lose friction with the track what is
called grip.
First we will attempt to fully understand acceleration and the relationships that define it
and velocity. Speed is the amount of distance covered in a period of time as noted in
equation (7.1). Velocity is a combination of speed and direction of the object. The speed
part of the relationship is defined by the following:
v=
d f − di
t f − ti
=
∆d
∆t
(7.3)
The df is final distance and di is the initial distance, this difference in distance is
represented by ∆d. The tf represents the final time and ti represents the initial time. The
difference in time is represented by ∆t. The slope of a line on a distance verses time
graph is the velocity. The velocity must have a direction such as 100 km/hr due east.
Many times the initial time is zero. This simplifies the relationship as the following:
d − di
v= f
tf −0
v=
d f − di
tf
t = tf
v=
d f − di
t
vt = d f − di
d f = vt + di
(7.4)
This is the typical form that we use to discuss slope (y = mx + b). Distance is plotted on
the y axis and time on the x axis.
If the velocity is not constant (linear and in a straight-line) then we have an acceleration.
If you refer to Figure 7.2 for the race car you see different shaped lines from horizontal to
diagonal to curves. The curved sections do not have a constant slope therefore they do
not have a constant velocity and are places during which acceleration is occurring. The
acceleration will always be a constant for any segment of the motion (zero acceleration is
still a constant). We plotted a graph of velocity vs time using the data in the table on
page one (calculations were done to determine instantaneous velocities).
90
80
70
velocity (m/s)
60
50
40
30
20
10
0
0
50
100
150
200
250
300
time (s)
Figure 7.3 Velocity vs Time Graph
The horizontal portions are when the race car is traveling at a constant velocity and from
the graph you can determine these velocities. The fifth segment is when the car is
stopped which is a constant velocity of zero. The diagonal lines of segments 2, 4, and 6
are linear and a slope can be determined. The slope of these diagonal lines represents the
acceleration during this period of time.
350
a=
v f − vi
t f − ti
ti = 0
tf = t
a=
(7.5)
v f − vi
t
at = v f − vi
v f = vi + at
This is one of the three equations of accelerated motion used to solve problems in this
area of physics. The three equations can be derived using algebra. The three equations
are:
v f = vi + at
(7.6)
1
d f − di = vi t + at 2
2
1
d f = vi t + at 2
2
2
2
v f = vi + 2a ( d f − di )
(7.7)
(7.8)
v 2f = vi2 + 2ad f
Note that in equations (7.7) and (7.8), di is normally equal to zero. The simpler form is
also shown and will be used. There are five different quantities (in the simplified form)
in equations (7.6), (7.7) and (7.8), knowing any three of these quantities gives us the
ability to solve for the other two quantities, assuming that di is zero.
EXAMPLE: A drag racer covers 400 meters in 5 seconds. The race car starts from rest.
What is the acceleration of the car?
First we must determine what is known in the problem. I suggest you make a table.
Quantity
Value
Units
vi
vf
t
df
a
0
m/s
m/s
s
m
m/s2
5
400
Next we must look at the three accelerated motion equations and determine for which one
we know three of the four variables. Hopefully it will contain the variable we wish to
determine.
In equation (7.6) we know the initial velocity and the time.
In equation (7.7) we know the distance (remember the initial distance is zero), we know
the initial velocity and the time. It also contains the variable of acceleration which we
want to determine.
Therefore we will use equation (7.7)
1
d f − di = vi t + at 2
2
1
d = vi t + at 2
2
1
m
2
400m = 0
( 5sec ) + a ( 5sec )
sec
2
2
400m = 0m + 12.5sec a
400m
a=
12.5sec 2
m
a = 32 2
sec
(7.9)
Now that we know the acceleration we can also determine the velocity of the car at the
end of the run. We must again determine which equation to use and we now also know
an additional variable, the acceleration (substitute this back into your table).
In equation (7.6) we know the initial velocity, the acceleration and the time. The
equation also has the unknown that we wish to solve, for the final velocity. If we use this
equation we get the following:
v f = vi + at
m
m
+ 32 2 ( 5sec )
sec
sec
m
v f = 160
sec
vf = 0
(7.10)
EXAMPLE: If a ball is thrown upward with a velocity of 20 meters per second how
long does it take to reach maximum height? The acceleration of gravity is a constant of
m
9.8 2 . When the ball is at maximum height it is not moving because in the next instant
s
in time it will begin to fall downward.
First we create a data table.
Quantity
Value
Units
vi
vf
t
df
a
20
0
m/s
m/s
s
m
m/s2
9.8
For equation (7.6) we know the initial and final velocity and the acceleration of gravity.
Therefore we can solve for the time, but this is not the quantity we are looking for. If
none of the other equations gives a direct solution we may need to use this equation. The
next equation (7.7) also has a time measurement in it and thus we will not initially use it.
Equation (7.8) has both the initial and final velocity and the acceleration of gravity. It
also contains the final distance which for this example would be the maximum height.
Using this equation we will solve the problem. A negative sign will be placed in front of
the acceleration of gravity because the ball is moving in an upward direction and the
acceleration of gravity is in the opposite direction (remember that acceleration and
velocity are vector quantities).
v 2f = vi2 + 2ad f
2
2
m⎞
⎛ m⎞ ⎛ m⎞
⎛
⎜ 0 ⎟ = ⎜ 20 ⎟ + 2 ⎜ −9.8 2 ⎟ d f
s⎠
s ⎠
⎝ s⎠ ⎝
⎝
2
2
m
m ⎛
m⎞
0 2 = 400 2 − ⎜19.6 2 ⎟ d f
s
s
s ⎠
⎝
m2
m⎞
⎛
= − ⎜ 19.6 2 ⎟ d f
2
s
s ⎠
⎝
2
m
−400 2
s =d
f
m⎞
⎛
− ⎜19.6 2 ⎟
s ⎠
⎝
d f = 20.4m
−400
(7.11)
Force
As previously noted if we have a change in velocity then we must also have acceleration.
This change in velocity might occur from a change in speed or a change in direction. A
race car leaving the pits will accelerate in a linear fashion, similarly as the car goes
around the corner on the race track it will accelerate. According to Newton’s Second
Law of Motion if there is acceleration then there must be an unbalanced force.
This force is defined as:
F = ma
(7.12)
F is the force measured in Newtons. m is the mass measured in kilograms, and a is the
acceleration measured in m/s2 . Return to the drag racer example which we solved for the
acceleration of the car. If the car has a mass of 750 kilograms we can now determine the
amount of force that must be applied between the tires and the pavement to cause the car
to accelerate down the track. Using Equation (7.12):
F = ma
⎛ m⎞
F = ( 750kg ) ⎜ 32 2 ⎟
⎝ s ⎠
F = 24000 N
(7.13)
Note that the unit of force in the SI system of measurement is the Newton, and a Newton
is equal to a kgm/s2.
When the race car goes around a corner we must use an equation derived from Newton’s
Second Law of Motion. We will not derive the equation in this discussion. The equation
is for centripetal force. Centripetal force is an inward force which is required for an
object to travel in a curved path. The force is always toward the center of the curve.
When you go around a curve you feel as though you are being pushed to the outside of
the curve because your body wants to continue in a straight line.
F⊥ = m
v2
r
(7.14)
Using the race car in the graph and using the last segment of the graph we can determine
that the car was traveling at 50 m/s. The car has a mass of approximately 1500 kg. We
must also know the radius of the curve on a race track. We will make an assumption for
simplification at this stage that the curve is flat (no banking) and is one quarter of a mile
in circumference. We must determine the length in meters and the radius of the curve. A
quarter of a mile is approximately 400 meters. Since this curve represents one quarter of
a circle we can use the equation for the circumference of a circle to determine the radius.
c = 4(400m)
c = 2π r
1600m = 2π r
1600m
r=
2π
r = 254.6m
(7.15)
Using equation (7.14) we can determine the force required for the car to go around this
curve.
F⊥ = m
v2
r
m ⎞
⎛
⎜ 50
⎟
sec ⎠
F⊥ = (1500kg ) ⎝
254.6m
F⊥ = 14729 N
2
(7.16)
This is the force required for the car to go around the curve. Without this force the car
will not travel around the curve and will strike the wall. This exemplifies Newton’s First
Law of Motion that an object travels in a straight line unless acted upon by an outside
force. The force required is from the tires in contact with the pavement. The force must
be inward toward the center of the curve along the radius. One of the items on which a
race team works very hard is to insure that the car can go around the curve easily but
have no more frictional force than is required because this will slow the car. This is done
in several ways. One way is putting different air pressure in the tires on the outside of the
car compared to those on the inside. This actually slightly changes the diameter of the
tire.
Banked Curve
If the car is not moving and sitting on the banked curve then the force of friction between
the tires of the car is equal to a component of the weight of the car. If the frictional force
was not large enough then the car would slide down the track. The frictional force can
never exceed the component of weight since it is a reactionary force. If all the forces are
balanced then we state that the object (car) is in equilibrium and if it is in equilibrium
then the first law of motion must apply since the net force is equal to zero.
When a race car goes around a banked curve the tires provide only a part of the
centripetal force; the weight of the car
on the banked curve applies the
remainder of the force. The diagram to
the left will assist in the understanding
of Newton’s Third Law of Motion. This
diagram is a cross section of a banked
curve on a race track, and the block
represents the car on the track. The third
law of motion states that for every action
there is an equal and opposite reaction.
The car pushes down on the track (a
component of the weight of the car) and
the track pushes back on the car. If this
Figure 7.5 Banked Curve
did not occur the car would sink into the pavement.
The weight of the car is represented with W. N is the Normal force which is the
pavement supporting the car. F is a component of the Normal force which assists the car
going around the curve. Remember unless this track has no friction this will only be one
of the two forces required for the car to go around the track, the other being a component
of friction between the tires and the pavement. To solve this problem we would need to
use either an accurate vector diagram (and do a graphical representation of the forces)
that represents the forces present or we need to use trigonometry. An understanding of
the banking of the curve is important but we will not attempt to solve this type of
problem in this chapter.