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equation of the form
ax2 + bx + c = 0
with the solution
-b ± b2 - 4ac
x =
2a
In this problem, x = t , a = 4.90, b = -17.5, and c = -30.0. Therefore,
t
2
+17.5 ± 17.5 - 4 4.90
( ) ( )(-30.0
)
tt =
2(4.90)
= +17.5 ± 29.9
9.80
= 4.84 s
The total time that the projectile is in the air is 4.84 s. If we had solved the equation for the negative root, we
would have found a time of -1.27 s. This corresponds to a time when the height is -30.0 meters but it is a time
before the projectile was thrown. If the projectile had been thrown from the ground it would have taken 1.27
seconds to reach the roof.
d. To find the velocity of the projectile at the ground we use equations 3.50, 3.51, and 3.53:
v = v = 30.3 m/s
xg 0x
v = v - gt = 17.5 m/s - (9.80 m/s2)(4.84 s)
yg 0y t
= -29.9 m/s
2 2
vg = vxg + vyg
( ) ( )
22
vg = 30.3 m/s +
() (-29.9 m/s
)
= 42.6 m/s
The speed of the projectile as it strikes the ground is 42.6 m/s. The angle
that the velocity vector makes with the ground, found from figure 3.29,
is
tan Æ = v
yg
v
xg
vyg
-29.9
Õ = tan-1 = tan-1 ëø öø
ìø ÷ø
vxg 30.3
íø øø
= -44.60
Figure 3.29 Angle of velocity vector as it
strikes the ground.
3-32 Mechanics
The velocity vector makes an angle of 44.60 below the horizontal when the projectile hits the ground.
e. To find the range of the projectile we use equation 3.46:
x = v t = (30.3 m/s)(4.84 s)
max 0x t
= 147 m
To go to this interactive example click on this sentence.
Have You Ever Wondered . . . ?
An Essay on the Application of Physics
Kinematics and Traffic Congestion
Have you ever wondered, while sitting in heavy traffic
on the expressway, as shown in figure 1, why there is
so much traffic congestion? The local radio station
tells you there are no accidents on the road, the traffic
is heavy because of volume. What does that mean?
Why can t cars move freely on the expressway? Why
call it an expressway, if you have to move so slowly?
Let us apply some physics to the problem to
help understand it. In particular, we will make a
simplified model to help analyze the traffic congestion.
In this model, we assume that the total length of the
expressway L is 10,000 ft2 (approximately two miles),
Figure 1 Does your highway look like this?
the length of the car x is 10 ft, and the speed of the car v is 55 mph. How many cars of this size can safely fit on
c 0
this expressway if they are all moving at 55 mph?
First, we need to determine the safe distance required for each car. If the car is moving at 55 mph (80.7
ft/s), and the car is capable of decelerating at -18.0 ft/s2, the distance required to stop the car is found from
equation 3.16,
v2 = v + 2ax
02
by noting that v = 0 when the car comes to a stop. Solving for the distance x that the car moves while decelerating
d
to a stop we get
x = -v = -(80.7 ft/s)2 = 181 ft
d 02
2a 2(-18 ft/s2)
Before the actual deceleration, the car will move, during the reaction time, a distance x given by
R
x = v t = (80.7 ft/s)(0.500 s) = 40.4 ft
R 0 R
where we assume that it takes the driver 0.500 s to react. The total distance "L needed for each car on the
expressway to safely come to rest is equal to the sum of the distance taken up by the car itself x , the distance the
c
car moves during the drivers reaction time x , and the distance the car moves while it is decelerating x . That is
R d
"L = x + x + x = 10 ft + 40.4 ft + 181 ft = 231 ft
c R d
2
We will depart from our custom of using only SI units here because most students will have a better feel for this discussion if it is done in the
British engineering system of units.
Chapter 3 Kinematics - The Study of Motion 3-33
Because it takes a safe distance "L for one car to come to rest, N cars will take a distance of N"L. The total
length of the road L can then hold N cars, each requiring a distance "L to stop, as seen in figure 2. Stated
mathematically this is
L = N"L (H3.1)
Figure 2 The number of cars on an expressway.
Therefore, the number of cars N that can safely fit on this road is
N = L = 10,000 ft = 43 cars
"L 231 ft
Thus for a road 10,000 ft long, only 43 cars can fit safely on it when each is moving at 55 mph. If the
number of cars on the road doubles, then the safe distance per car "L must be halved because the product of N and
"L must equal L the total length of the road, which is a constant. Rewriting equation H3.1 in the form
"L = L
N
= x + x + x = L
c R d
N
x + v t - v = L (H3.2)
c 0 R 02
2a N
Notice in equation H3.2 that if the number of cars N increases, the only thing that can change on this fixed length
road is the initial velocity v of each car. That is, by increasing the number of cars on the road, the velocity of each
car must decrease, in order for each car to move safely. Equation H3.2 can be written in the quadratic form
- v + v t + x - L = 0
02 0 R c
2a N
which can be solved quadratically to yield
2
v0 = atR m atR + 2a xc - L / N (H3.3)
( ) ( )
Equation H3.3 gives the maximum velocity that N cars can
safely travel on a road L ft long. (Don t forget that a is a
negative number.) Using the same numerical values of a, t ,
R
x , and L as above, equation H3.3 is plotted in figure 3 to
c
show the safe velocity (in miles per hour) for cars on an
expressway as a function of the number of cars on that
expressway. Notice from the form of the curve that as the
number of cars increases, the safe velocity decreases. As the
graph shows, increasing the number of cars on the road to
Figure 3 Plot of the velocity of cars (y-axis) as function
of the number of cars on the expressway (x-axis).
3-34 Mechanics
80, decreases the safe velocity to 38 mph. A further increase in the number of cars on the road to 200, decreases
the safe velocity to 20 mph.
Hence, when that radio announcer says,  There is no accident on the road, the heavy traffic comes from
volume, he means that by increasing the number of cars on the road, the safe velocity of each car must decrease.
You might wonder if there is some optimum number of cars that a road can handle safely. We can define
the capacity C of a road as the number of cars that pass a particular place per unit time. Stated mathematically,
this is
C = N (H3.4)
t
From the definition of velocity, the time for N cars to pass through a distance L, when moving at the velocity v , is
t = L
v
Substituting this into equation H3.4 gives
C = N = N = v
t L/v L/N
Substituting from equation H3.2 for L/N, the capacity of the road is
C = v (H3.5)
x + v t - v /2a
c 0 R 02
Using the same values for x , t , and a as before, equation H3.5 is plotted in figure 4. The number of cars per hour
c R
that the road can hold is on the y-axis, and the speed of the cars in miles per hour is on the x-axis. Notice that at a
speed of 60 mph, the road can handle 1200 cars per hour. By decreasing the speed of the cars, the number of cars
per hour that the road can handle increases. As shown in the figure, if the speed decreases to 40 mph the road can
handle about 1600 cars per hour. Notice that the curve peaks at a speed of about 13 mph, allowing about 2300
cars/hour to flow on the expressway. Thus, according to this model, the optimum speed to pass the greatest
number of cars per hour is only 13 mph. Hence, even though the road may be called an expressway, if the volume
of cars increases significantly, the cars are not going to travel very rapidly. The solution to the problem is to build [ Pobierz całość w formacie PDF ]

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