Solution to Assignment 3
PROBLEM 1 – 10 points
On their way to play soccer in the
World Cup, Cindy and Mia get
stranded in the Chicago airport
because of bad weather. Late at
night, with nobody else around,
they decide to have a race to prove
who is faster. They start at the same time, run at top speed a distance of L = 120 m through the
airport terminal, and then turn around and run back to the starting point. Cindy runs at a constant
speed of v. Mia runs on a moving sidewalk that travels at a speed of 4.00 m/s; she runs at a
constant speed of 8.00 m/s relative to the moving sidewalk. The race turns out to be a tie. For
your analysis below, neglect the time it takes the women to turn around.
(a) How long does it take Mia to reach the turn-around point? How long does it take her to get
from the turn-around point back to the start?
Soln: The first step in solving both the parts of this part of the problem is to find the speed of
Mia with respect to ground. The relation between the velocities of Mia with respect to sidewalk,
vMS, velocity of Mia with respect to ground vMG and the velocity of sidewalk with respect to the
ground vSG are given by
vMG = vMS + vSG.
When Mia is running towards the turn around point, she is running with the sidewalk. Therefore,
the speed of Mia with respect to ground will be (8 + 4 ) ms-1= 12 ms-1. Therefore,. She takes 10
seconds to reach the turnaround point.
When Mia turns around, she runs against the direction of motion of sidewalk. Therefore her
speed with respect to ground will be (8 – 4) ms-1 = 4 ms-1. Therefore, she will take 120m / 4ms-1
= 30 seconds to reach the starting point from the turn around point.
(b) What is Cindy’s constant speed v?
Since, the race ends up in a tie, we conclude that Cindy takes (10 + 30) s = 40 s to complete the
race. Since we know that Cindy’s speed is constant, her speed is given by 240m/40s = 6 ms-1.
(c) Who is the faster runner?
[x ] Mia
[ ] Cindy
[ ] Neither, they’re equally fast
Briefly justify your answer:
Mia can run at 8 ms-1 with respect to any surface whereas Cindy can run only at 6 ms-1 with
respect to any surface. Therefore, if Mia had run on the ground, instead of the sidewalk, she
would have won the race. Hence, the conclusion that Mia is the faster runner.
(d) At what time in the race is the distance between the women the largest? What is the distance
between them at this time?
As the race begins, Cindy runs at 6 ms-1 and Mia moves ahead towards the turn around point at
8ms-1. Therefore, the distance between Cindy and Mia increases till Mia reaches the turnaround
point. Then, Mia turns around and starts going towards the starting point approaching Cindy in
the opposite direction and thus the distance between Cindy and Mia starts decreasing. Therefore,
the distance between Mia and Cindy is greatest, when Mia is at the turn around point.
We know that Mia reaches turn around point at t = 10 seconds after the race begins and Cindy
would have travelled a distance of 6 ms-1 * 10 s = 60 m from the starting point. Therefore the
distance between Cindy and Mia is 120 m – 60m = 60 m. Therefore, the greatest separation
between Mia and Cindy during the entire race is 60m.
(e) In addition to the start and end of the race, at what other location(s) is/are the women the
same distance from the start/finish line at the same time? Express the location(s) in terms of the
distance from the start/finish line.
Consider the situation shown in the figure below. Mia turns around and starts approaching Cindy
from the opposite direction. Let the distance of Cindy from the starting point be ‘x’m, when
Cindy and Mia meet. Therefore, from the figure, we see that Mia would have travelled a distance
of (120-x )m. Therefore, the time taken by Cindy to cover ‘x’m is (x/6) seconds, which will be
the same time that Mia takes to go to the turn around point and then cover a distance of (120x)m after turning around.
Setting Cindy’s time equal to Mia’s time, we get,
x 120 120 − x
=
+
6 12
4
Simplifying the above expression, we get x = 96. Therefore, Cindy and Mia meet at 96m from
the starting point(apart from the beginning and end of the race).
PROBLEM 2 – 10 points
You are flying a small plane from Boston to Buffalo, which is located 660 km due west of
Boston. Immediately after taking off, you point your plane due west, set the autopilot to cruise at
a speed of 220 km/h relative to the air, and then you take a nap. You wake up later and, after
checking your watch, you expect to be directly over your destination (you assumed there was no
wind at all). Instead, you find yourself 240 km south and 480 km west of Boston, over
Harrisburg, Pennsylvania instead.
(a) Sketch a diagram, showing the locations of Boston, Buffalo, and Harrisburg.
The problem says that Buffalo is 660 km due west of Boston and Harrisburg is 240 km due south
and 480 due west of Boston. Therefore, the diagram is as shown in the figure.
(b) How long after taking off did you wake up from your nap?
The problem says that you wake up expecting to be over Buffalo. Therefore, if there was no
wind you would have covered 660 km at a speed of 220 km/h. Therefore, you would have set the
alarm after 660km/220 km/h = 3 h.
(c) Assuming the autopilot did exactly what you told it to do, what was the average velocity of
the wind acting on your airplane during the flight? You can express this in terms of its
components.
The equation that relates the velocity of plane with respect to ground, vPG , velocity of plane with
respect to air, vPA and the velocity of wind, vWG is
vPG = vPW + vWG
Taking positive x and y directions along North and East directions, we get the vPW=220 km/h 𝑥
and vPG= -160 km/h 𝑥 - 80 km/h y (The plane covers 480 km due west and 240 km due south in
three hours.) . Therefore, we get vWG= 60 km/h 𝒙 - 80 km/h 𝐲.
(d) When you realize that the wind has blown you off course, you immediately change direction
so that you will reach Buffalo. Accounting for the wind, do you point your plane directly toward
Buffalo, or not? Explain your answer.
The velocity of the wind makes and angle of tan−1 ( 80/60), clockwise with respect to positive
x-direction and Harrisburg is at an angle of tan−1 ( 240/180)= tan−1 ( 80/60). Therefore, wind
blows from Buffalo to Harrisburg.
If we point our direction of plane towards Buffalo, then we go directly against the wind and
hence we will not be deflected off. Hence the plane should be pointed towards Buffalo.
(e) Assuming your speed is still 220 km/h with respect to the air, how long does it take for you to
travel from Harrisburg to Buffalo? The wind is still blowing.
We know that when we point the plane towards Buffalo, the plane goes at 220km/h against the
wind. The velocity of the wind is 100km/h (can be found using Pythagorean theorem from the
components in part c) directed from Buffalo to Harrisburg. Therefore, the velocity of the plane
with respect to ground will be (220- 100= 120)km/h. Therefore the plane would take 300 km /
120km/h = 2.5 hours to reach Buffalo from Harrisburg.
PROBLEM 3 – 10 points
In solving physics problems, it can be helpful to use a systematic approach. In this problem, we
will work through an approach that you should be able to use for all projectile-motion problems.
Working as an accident reconstruction expert, you find that a car that was driven off a horizontal
road over the edge of a 15-m-high cliff traveled 45 m horizontally before impact.
Use g = 10 m/s2.
(a) Sketch a diagram of this situation, starting from where the car leaves the road to where it hits
the ground. On your diagram, show the origin, and the coordinate system (show the directions
you are taking for positive x and positive y).
(b) Fill in the table,
to help you stay
organized. This also
reminds you to keep
the x information
separate from the y
information when
you use the
equations.
𝐴 = 𝜋𝑟 2
x-direction
y-direction
displacement
(45 – 0)m = 45m
( 0 - 15)m = -15 m
initial velocity
We‟ll find this is (d)
Zero(because car
moves horizontally
before falling off.)
-10 m/s2
acceleration
Zero m/s2
Use the information
in the table to help answer the questions below.
(c) How long after leaving the road did the impact with the ground occur?
We know that the car falls through a distance of 15m in the vertical direction. Therefore, we can
Using the equation,
yf = yi + vi,y t - ½ g t2
and the information that yi = 15m , yf = 0 m and vi,y = 0 m/s, we get t = 𝟑𝒔 = 1.7 s.
(d) At what speed was the car moving when it left the road? Express the speed in m/s and km/h.
We know that the car cover a distance of 45 m in the horizontal distance in 3𝑠 and the
acceleration of car along horizontal direction is zero. Therefore the initial velocity of car, when
the car left the road is 45m/ 3𝑠 = 15 3𝑚/𝑠 = 26.0 m/s.
(e) At what speed was the car moving just before impact?
We know that the horizontal component of velocity doesn‟t change. The vertical component of
velocity of the car just before the impact is found using the relation(with t = 1.7s),
vf,y = vi,y - g t,
is found to be -10 3𝑠 = 17.3m/s. Therefore the magnitude of the velocity of the car before the
impact is found, using Pythagorean theorem to be 31.2 m/s.
PROBLEM 4 – 10 points
The motion diagram shows a ball‟s position at intervals of exactly 1.00 seconds as the ball flies
from left to right through the air, after being launched from the origin at t = 0. Only gravity acts
on the ball throughout the motion.
Use g = 10 m/s2.
(a) Calculate the y-component of the ball‟s initial velocity.
From the motion diagram we see that the ball reaches highest point at 4 seconds after it is
launched. Also, the vertical component of velocity is zero at the highest point. Therefore, we
find the y-component of initial velocity using the equation,
vf,y = vi,y - g t,
as + 40m/s.
(b) Calculate the maximum height reached by the ball.
The maximum height reached by the ball can be found using the relation,
yf = yi + vi,y t - ½ g t2
with t = 4s and vi,y= 40 m/s. The maximum height reached by the ball is found to be 80 m.
(c) Calculate the x-component of the ball‟s initial velocity.
Assuming that the scales of x and y axes to be same, we infer from the motion diagram that the
ball covers a distance of 80m, along horizontal direction, in 8 seconds. Therefore, the xcomponent of initial velocity of ball is found to be 80m/8s = 10m/s.
(d) On the graph above, sketch the motion diagram for a second ball that has the same starting
point, half the initial vertical velocity, and three times the horizontal velocity, of the original ball.
By “sketch a motion diagram,” we mean “show the position of this ball at 1.00-second intervals”.
The black ball in the figure above represents the position of the second ball.
(e) Calculate the maximum height reached by the second ball.
From the figure, we see that the second ball remains in flight for 4 seconds. Therefore the ball
reaches the maximum height at t = 2 seconds. Therefore, using the equation,
yf = yi + vi,y t - ½ g t2
with yi = 0m and vi,y = 20 m/s, we get the maximum height to be 20m.
PROBLEM 5 – 8 points
Three students are trying to solve a problem that involves a ball being launched, at a 30° angle
above the horizontal, from the top of a cliff, and landing on the flat ground some distance below.
The students know the launch speed, the acceleration due to gravity, and the height of the cliff.
They are looking for the time the ball spends in the air. A snippet of their conversation is
recorded below.
Avi : I think we have to do the problem in two steps. First, we find the point where the ball
reaches its maximum height, and then we go from that point down to the ground.
T.J. : I think we can do it all in one step. The equations can handle it, just going all the way
from the initial point to the ground.
Kristin: I think we need two steps, too, but I would do it differently than Avi. What if we first find
the point where the ball comes down to the same height from where it was launched, and then we
go from that point down to the ground?
(a) Can you do the problem in two steps, like Avi and Kristin suggest? If so, comment on which
approach, Avi or Kristin‟s, you prefer.
Yes. We can do the problem in two steps in both ways suggested by Avi and Kristin. There is no
difference in the amount of work each method takes. Therefore, there is no preference.
(b) Can you do the problem all in one step, like T.J. says? If so, is this preferred or not compared
to a two-step method?
Yes. We can do the problem in one step as TJ suggests. This method involves solving a quadratic
equation and it depends on each one‟s comfort with quadratic equation to decide about the
preference.
(c) If T.J. is correct, can you identify a single equation that can take all the known quantities, and
that you can solve for the time? If so, what equation would you use?
yf = yi + vi,y t - ½ g t2
and we know all the quantities in the above equation except time „t‟.
(d) Using a cliff height of 20 m, a launch speed of 8.0 m/s, and g = 10 m/s2, solve for the time of
flight.
The problem can be solved using the above equation with yi = 20m , yf = 0 m and vi,y = 8
sin(30)m/s = 4 m/s. Therefore, solving the quadratic equation for time we get the time that the
ball remains in air to be 2.4s.