PHYS 2210 — Fall 2015
GA1 Solutions
Use the general problem solving framework:
I. Understand the Problem
a) Read the problem carefully.
b) Construct a mental image.
c) Determine the question.
d) Summarize the given information.
II. Describe the Physics
a) Draw a useful diagram.
b) Assign symbols to known and unknown quantities.
c) Label relevant quantities on the diagram.
d) Declare the target variable.
e) State the relevant physical principles.
III. Plan the Solution
a) Write down relevant equations.
b) Write down relevant constraints.
c) State the necessary approximations.
d) Outline how to use the equations to determine the target variable.
IV. Execute the Solution
a) Solve for the target variable symbolically (algebraically, not numerically).
b) Check the units of the equation.
c) Substitute numerical values of known quantities.
d) Calculate the value of the target variable.
e) Answer the question.
V. Evaluate the Solution
a) Is the answer properly stated?
b) Is the answer reasonable?
c) Is the answer complete?
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2
Problem 1
One-Dimensional Kinematics: Bonnie and Clyde
In your new job, you are the technical advisor for the writers of a gangster movie about Bonnie
and Clyde. In one scene Bonnie and Clyde try to flee from one state to another. If they get across
the state line, they could evade capture, at least for a while until they become Federal fugitives.
In the script, Bonnie is driving down the highway at 108 km/hr and passes a concealed police car
that is 1 km from the state line. The instant Bonnie and Clyde pass the patrol car, the cop pulls
onto the highway and accelerates at a constant rate of 2 m/s2 . The writers want to know if they
make it across the state line before the pursuing cop catches up with them.
Solution
I. Understand the Problem:
Since Bonnie and Clyde are traveling at a constant velocity and the cop is accelerating, the cop
will eventually overtake Bonnie and Clyde at a certain place and time. Our job is to determine
whether this place is before or after the state line, or equivalently, whether this time is shorter or
longer than the time taken for Bonnie and Clyde to reach the state line.
II. Describe the Physics:
Always start with a diagram and a coordinate system. In this case, we use a 1D coordinate system
where x is the direction of travel for the two cars and their initial position is at x = 0. The diagram
denotes the initial situation when Bonnie and Clyde race past the cop.
Bonnie and Clyde:
• The position of Bonnie’s car when she passes the stationary cop is xB,i = 0.
• Bonnie’s initial velocity is toward the right with speed vB,i = 108 km/h.
• Bonnie maintains a constant velocity, so her acceleration is zero: aB = 0.
• Bonnie’s position when her car is overtaken by the cop is: xB,f = xf . We don’t know this
(we’ll solve for it), but if xf < 1 km, Bonnie and Clyde are caught. If xf > 1 km, Bonnie
and Clyde escape.
Cop:
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• The initial position of the cop’s car is xC,i = 0.
• The cop’s initial velocity is zero: vC,i = 0.
• The cop accelerates to the right with constant acceleration: ac = 2 m/s2 .
• The cop’s position when he overtakes Bonnie and Clyde is: xC,f = xf . We set this distance
equal to the “final” position of Bonnie and Clyde - that is, the distance they travel before
being overtaken by the cop.
We are going to use the 1D kinematics equations with constant acceleration since both cars have
constant acceleration: aB = 0 (0 is a perfectly legitimate constant value), and aC = 2 m/s2 .
III. Plan the Solution
Each car needs an equation to describe its position as a function of time and each equation can
only describe a single car:
=⇒ We’ll have two equations.
Which equations should we choose? Since we’re interested in both cars’ final position (which will
be the same value, xf when the cop overtakes Bonnie and Clyde), we might guess that each car’s
position would be described by the following two equations:
1
xf = xi + vi t + at2 , or
2
vf2 = vi2 + 2a(xf − xi ).
(1)
(2)
where t is the amount of time between when Bonnie and Clyde pass the stationary cop, and when
the cop catches up. For Bonnie and Clyde, the second equation only confirms that vf = vi (i.e.,
their speed is constant) since aB = 0. Furthermore, for the Cop, we don’t know either the time
t it takes for him to pass Bonnie and Clyde, nor the Cop’s final velocity, vC,f , but DO know
that Bonnie and Clyde and the cop travel the same distance in the same amount of time. Thus, it
makes sense to use the first general equation above both for Bonnie and the cop, with each equation
written with the particular variables corresponding to that particular car.
IV. Execute the Solution
Now that we’ve decided to use the first equation above for both cars, we should write two equations
with the appropriate variables:
1
xf = xB,i + vB,i t + aB t2 = vB,i t (Bonnie’s initial position and acceleration are both zero) (3)
2
1
1
xf = xC,i + vC,i t + aC t2 = aC t2 (the cop’s initial position and velocity are both zero).
(4)
2
2
Notice that these two equations describe the motion of the car (after passing the cop) and the
cop (after being passed), until the cop catches up. Also notice, that between these two equations,
we only have two unknowns: t and xf . We know that with two equations, we can solve for both
unknowns. The one we care about the most is xf , because if xf < 1 km, the cop catches Bonnie
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and Clyde before they cross the state line. A simple strategy for finding xf is to solve the first
(simpler) equation for t and then substitute that into the second equation:
xf
vB,i
1
1
=⇒ xf = aC t2 = aC (xf /vB,i )2
2
2
2
vB,i
.
=⇒ xf = 2
aC
xf = vB,i t =⇒ t =
(5)
(6)
(7)
We’ve been given the values for both vB,i (100 m/s) and aC (2 m/s2 ), so if Eqn. (7) is correct,
we should be able to calculate xf and answer the question. Before doing that, however, let’s check
whether the expression makes sense by doing a little dimensional analysis. The units of speed for
Bonnie and Clyde (vB,i ) are given in km/h, or [length/time]. The units of acceleration for the cop
(aC ) are given in m/s2 , or [length/time2 ]. So the right side of Eqn. (7) will have units of:
[length2 ]
[time2 ]
[length]
[time2 ]
= [length],
(8)
which is correct. (This doesn’t prove that Eqn. (7) is correct - the factor of 2 could be wrong, for
example - but at least it’s reasonable!) Now, we want to solve Eqn. (7) numerically, and get a
number for xf so we can answer the question: Does the cop catch Bonnie and Clyde before they
cross the state line? It will help if we first convert Bonnie and Clyde’s speed from km/h to m/s so
that everything will be in Standard Units (SI):
km
1000 m
1000 m
1
=
=⇒ 108 km/h = 108
(9)
h
3600 s
3600 s
= 30 m/s.
(10)
So now, we find a numerical solution for Eqn. (7):
xf = 2
2
vB,i
aC
(30 m/s)2
2 m/s2
= 900 m.
=2
(11)
(12)
V. Evaluate the Solution
Since xf = 900 m, the cop catches Bonnie and Clyde before they cross the state line (1 km from
the initial position of the cop)! Note that if xf had turned out to be larger than 1 km, then Bonnie
and Clyde would have escaped.
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Problem 2
One-Dimensional Kinematics: Catching the Train
You are going to Chicago for the weekend and you decide to go first-class by taking the AmTrak
train. Unfortunately, you are late finishing your mathematics exam, so you arrive late at the train
station. You run as fast as you can, but just as you reach the platform your train departs, 30
meters ahead of you down the platform. You can run at a maximum speed of 8 m/s and the train
is accelerating at 0.8 m/s2 . You can run along the platform for 50 meters before a barrier prevents
you from going further. Will you catch your train?
Solution
I. Understand the Problem:
According to the problem, when you arrive at the platform, the train (presumably the back of the
train) is already ahead of you by 30 m - its initial position is 30 m larger than your initial position.
You are running at full speed with a speed of 8 m/s . You maintain this speed while the train
accelerates from rest at a rate of 0.8 m/s2 . The train begins moving at the same moment you arrive
at the platform. There is a barrier preventing you from running more than 50 meters from your
initial position in the direction of the train. Since the train starts out already 30 m ahead, you
must catch it before it moves an additional 20 m.
II. Describe the Physics:
Once again, we start with a diagram and a 1D coordinate system where the train and the runner
move in the positive x direction. The diagram denotes the situation when the runner enters the
platform: she can see the back of the train 30 meters ahead and a barrier 50 meters ahead.
Runner:
• The runner’s initial position when she enters the platform is xR,i = 0.
• The runner’s initial velocity is to the right with a speed of vR,i = 8 m/s. She doesn’t change
her speed, so
• The runner’s acceleration is aR = 0.
• The runner’s final position (when she catches up to the train) is xf . We must solve for this
variable, but we already know that if xf < 50 m, she will catch the train, while if xf ≥ 50 m,
she will hit the barrier and thus won’t catch the train.
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Train:
• The train’s initial position (when the runner enters the platform) is xT,i = 30 m. This is the
position of the rear doors of the train.
• The train’s initial velocity is vT,i = 0.
• The train’s acceleration is to the right with a value of aT = 0.8 m/s2 .
• The final position of the train’s rear doors (when the runner catches up) is xf . Again, if xf
is larger than 50 m, the runner will hit the barrier before catching the train.
We will utilize the 1D kinematic equations since both the train (aT = 0.8 m/s2 ) and the runner
(aR = 0) have constant acceleration for the duration of the problem.
III. Plan the Solution:
We need one equation for the train and one for the runner. In this case, the initial positions of the
train and runner are not the same (as they were in the Bonnie and Clyde problem above), but we
can still utilize the same general 1D kinematic equation for constant acceleration for both:
1
xf = xi + vi t + at2 ,
2
(13)
where once again, t is the time it takes for the runner to catch the train at xf . In the case of the
train, we will substitute xi = xT,i = 30 m, vi = vT,i = 0, and a = aT = 0.8 m/s2 . For the runner,
we have xi = xR,i = 0, vi = vR,i = 8 m/s, and a = aR = 0. As before, we will write down two
equations with the above substitutions and then solve for xf .
IV. Execute the Solution:
The two 1D kinematic equations describing the motion of the train and runner are:
1
1
xf = xT,i + vT,i t + aT t2 = xT,i + aT t2 (Train’s motion.)
2
2
1
2
xf = xR,i + vR,i t + aR t = vR,i t (Runner’s motion.)
2
(14)
(15)
Again, let’s eliminate t by solving the second equation for t and substituting into the first equation:
xf = vR,i t =⇒ t =
xf
vR,i
x2f
1
1
=⇒ xf = xT,i + aT t2 = xT,i + aT 2
2
2 vR,i
aT 2
=⇒ 2 xf − xf + xT,i = 0.
2vR,i
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(16)
(17)
(18)
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This is a quadratic equation of the form: ax2 + bx + c = 0, where a =
aT
2 ,
2vR,i
b = −1, and c = xT,i .
To solve it, we use the quadratic formala:
x=
√
b2 − 4ac
s 2a aT
(xT,i )
1 ± 1 − 4 2v2
−b ±
(19)
R,i
=⇒ xf =
aT
2
vR,i
1±
=
q
a x
1 − 2 Tv2 T,i
R,i
aT
2
vR,i
(20)
(21)
2 has units of 1/[length], so a x /v 2 has no units
Dimensional Analysis: The quantity aT /vR,i
T T,i R,i
(dimensionless). Thus, Eqn. (21) has units of [length]. So at least Eqn. (21) is reasonable.
2 for the values given (a = 0.8 m/s2 ; v
Numerical Analysis: Let’s calculate aT /vR,i
T
R,i = 8 m/s):
aT
0.8 m/s2
0.8 −1
=
=
m
2
2
64
vR,i
(8 m/s)
= 0.0125 m−1 .
(22)
(23)
So Eqn. (21) becomes:
1±
q
a x
1 − 2 Tv2 T,i
R,i
xf =
aT
2
vR,i
p
1 − 2(0.0125 m−1 )(30 m)
=
0.0125 m−1
√
1 ± 1 − 0.75
1 ± 0.5
=
=
−1
0.0125 m
0.0125 m−1
= 40 m or 120 m.
1±
(24)
(25)
(26)
(27)
V. Evaluate the Solution:
The quadratic equation (21) has two solutions, one of which (xf = 40 m) is smaller than 50 m,
where the barrier is located. Thus, the runner will catch her train with 10 meters to spare! What
does the second solution, xf = 120 m, mean? Well, if the barrier were not present, and the runner
continued at a speed of 8 m/s instead of jumping into the train, then the runner would initially
pass the back doors of the train at 40 m, and then the train would catch up and pass the runner
(for the last time) at 120 m. Thus, without the barrier, the runner would get two chances to enter
the train. This is represented by the graph below.
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Distance (m)
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Time (s)
Figure 1: The solid line represents the motion of the train (quadratic time dependence) and the
dashed line represents the motion of the runner (linear time dependence). The runner will catch
the train after 5 s; 40 m from her starting position (10 m before the barrier). In the absence of the
barrier, the train would catch up to the runner at 120 m.
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