Chapter 3- Two-Dimensional KINEMATICS Hints –Updated 9/12/11
1. For the following cases of two-dimensional motion draw motion diagrams showing the velocity
vectors every second, the change in velocity vectors and the acceleration vectors for each second
interval using an arbitrary scale. The illustrations in the chart show the object in motion at each
second.
Velocity motion
diagram
Change in velocity
diagram
Acceleration diagram
A projectile rises for
2 sec and then falls
for 3 sec. Neglect
air resistance.
A rider at the edge
of a merry-go-round
rotating one
revolution in 6 sec
with constant speed.
A rider at the edge
of a merry-go-round
rotating one
revolution in 6 s
with increasing
speeds.
A pendulum swings
down for 2 s and up
2 s.
2. For a projectile with negligible air resistance (as in the question above) draw graphs of the horizontal
motion (x, vx, & ax vs. time). Also draw graphs of the vertical motion (y, vy, & ay vs. time). Finally draw a
graph of “y vs. x”. Which graph represents what you actually see when you watch the projectile in motion?
a) Is there a point in this motion where the velocity or the acceleration is zero? … or where one
component of the velocity or acceleration is zero?
b) Is there a point in this motion where the velocity and the acceleration are perpendicular to each
other?…or where they are parallel to each other?
c) On the same graphs you have already drawn indicate qualitatively how the curves would differ if air
resistance were not negligible. Recall that air resistance depends on the squared speed of the moving
object and acts always in opposition to the motion.
Assuming no air resistance, that the – negative direction is downward, and object lands below its origin:
x vs. t
vx vs t
ax vs t (no acceleration)
y vs. t
vy vs t
ay vs t (a=-g)
y vs. x ( This is the graph that resembles “reality” the most.) Solid line is the frictionless case and broked line
is roughly the air resistance case.
a) The acceleration is never zero (a= “g” always). The velocity is never zero either, but the y-component is
zero at the top of the trajectory.
b) At the top of the trajectory, v is horizontal and a is vertical.
c) Air resistance increases the upward deceleration and decreases the downward acceleration due to gravity.
It also decelerates the x-velocity. The broken line in the graph above is an approximation. The actual curve
is not a parabola, and it is not symmetric about the top position.
3. A particle moves along a path illustrated below. From A to B the path is curved, from B to D the path is
straight, from D to E the path curves the opposite way. (a) Given the velocity vector at A draw
qualitatively appropriate velocity vectors for all the other points in the following cases: (I) the particle
moves with constant speed, (II) the particle moves with increasing speed and (III) the particle moves with
decreasing speed. Also draw appropriate average acceleration vectors from A to B, from B to D, and from
D to E. Roughly…solid points are v-vectors and non-solid points are a-vectors:
(I)
(II)
(III)
E
E
E
C
C
D
B
D
C
B
D
B
A
A
A
(b) Let’s say that in (II) the particle has an initial velocity in the y-direction at A of 5 m/s. From A to B a
constant acceleration is applied in the x-direction for 3 seconds and the particle moves equal distances in
the x and y directions during that time. Determine the value of the acceleration.
The value of the acceleration in II from A to B is 10/3 m/s2. Given that the acceleration is restricted to the
horizontal direction, x=at2/2. In the y-direction, the velocity is constant, so y=vyt. Use the given
information: t=3; vy=5; x =y; to find the answer.
4. A plane is flying at an altitude of 300 m with a horizontal speed of 60 m/s
when it releases a package aimed at a target on the ground. Ignore air resistance.
Target
a) Where does the package land compared to the initial position of the plane?
b) If the plane doesn’t change its velocity, how far is it from the package when it hits the ground?
c) At what horizontal distance from the target must the pilot release the package in order to ensure that it
will reach its target?
d) The distance determined in (c) is not something that a pilot can easily determine while in flight, instead
he must sight the target and release the package at the proper “angle of sight” (denoted by the broken
line in the illustration) in order to hit the target. What is the appropriate angle of sight in this case?
Compare this to the “angle of impact” that the package makes with the ground when it lands.
e) To generalize this problem, assume that the height of the plane is H, its horizontal speed is vo, and the
acceleration of gravity is g. Derive the formula for the ”angle of sight” in terms of H, vo, and g.
a) 300 m below and 465 m ahead of starting point.
b) 300 m above
c) 465 m ahead
d) angle of sight= 33o with the horizontal, angle of impact= 52o with the horizontal
e) tan  = (Hg/2)1/2(vo)-1
5. A ball is fired with a velocity 5 m/s at an angle above the horizontal of The ground lies 2 m below the
point of release of the ball.
a) Determine the total time the ball spends in flight. Also determine the time it takes the ball to get to its
maximum height, and the time for it to be 2 m above the ground when coming down.
b) What is the maximum altitude reached by the ball and what is its overall range.
c) Determine the final velocity of the projectile when it hits the ground.
d) Redo the problem assuming that the initial angle was below the horizontal.
Some answers: (a) time in air= 1 sec; (b) H= 2.45 m;
(d) time in the air= 0.4 sec; range= 1.6 m; final vel=5.7 m/s at 45o
6. Projectile problems are all solved basically the same way: You sort out the given information into x&y
components, then you can analyze the information in each direction separately, using the kinematics formulas
for constant acceleration (if air resistance is negligible the only acceleration is “g” in the y-direction). Finally,
you solve for the unknowns. The problems, however, can be easier or harder depending on which variables are
“given” and which are the “unknowns”. Consider the following rewrites of the previous problem.
a) Redo the problem assuming that you are given the initial height “h”, the initial speed “vo”, and the time
in flight “t”. Determine the initial direction of the projectile and the overall range.
b) Redo the problem assuming that you know the initial height, the range, and the initial direction of the
projectile. Determine the initial speed and the time in flight.
The point here is to give you practice doing basic projectile problems. Some “givens” and “unknowns” are
“switched”. Hints:
a) The direction angle can be determine from the vertical displacement, g, and time in the air.
b) Combine the vertical and horizontal displacement formulas to eliminate one of the unknown variables.
7. In class we showed that a projectile landing at the same level from where it was launched has the maximum
range when the launch angle is 45o. This is not true, however if the landing place is at a different level from the
launch. Consider problem 6 above.
a) Show that changing the launch angle to 45o (from 37o) actually decreases the range in problem 6.
b) A mathematical derivation of the generalized maximum-range formula is quite difficult, but can you
give a conceptual explanation for the results in (a)?
c) To maximize the range of a projectile that lands at a higher level from where it was launched, would
one need to launch the projectile at an angle greater than 45o of less than 45o? Give a conceptual
argument to justify your answer.
a) The range is 3.8 m at the 45º-launch angle, which is less than the 4 m range at the 37º-launch angle.
b) When a projectile lands are a lower level, a smaller-than-45º launch angle decreases the time in free-fall
but increases “vx” by a larger factor. Therefore the range is maximized.
c) Launch angle should be greater than 45º. For a projectile that lands are a higher level, a larger-than-45º
launch angle decreases “vx” but increases the time in free-fall by a larger factor. This is difficult to prove
mathematically.
8. An athlete “launches” himself or herself into the air by pushing hard against the ground. In this problem
we compare the height of the “high jump” to the range of the “long jump”, assuming the initial speed of the
athlete, vo, is the same in both cases. Ignore air resistance and let the acceleration due to gravity be “g”.
a) Determine the height of a straight, vertical jump (“high-jump”) in terms of vo and g.
b) Determine the range of a horizontal jump on level ground (“long-jump”) in terms of vo and g. Use the
fact that the optimum launch angle for maximum range is 45o. Compare to the “high-jump”.
c) In the 1990’s the record “high jump” was set at about 2.45 m and the record “long jump” at about 8.3
m. Do these records support your solution to (a) and (b)? Are there some incorrect assumptions that
were made that could account for the discrepancy between the facts and the theory?
d) In the moon the acceleration due to gravity is g/6. How would that affect a “high-jump” and “longjump” in the moon?
a) H= vo2/2g ; b) max R= vo2/g
c) No, the solution to (a) & (b) suggest that the “long-jump” range should be twice larger than the “highjump” height, but these records show a ratio over three times larger. It suggests that the initial speed in the
“high jump” is actually less than in the “long-jump”. The initial velocities in the long-jump are about 2.5
times higher than in the high-jump, which is reasonable since an athlete doesn’t have to change his direction
of motion as much in the long-jump.
e) The formulas above are affected by the same 1/6 factor. The jumps would be 6 times greater in the moon.
9. A rocket is being fired from the ground at angle of 53o. While the rocket fuel lasts the rocket accelerates
uniformly for 4 sec and reaches a speed of 100 m/s starting from rest. Ignore air resistance.
a) Determine the maximum altitude that the rocket will
attain before it falls back to the ground.
b) Determine the total time the rocket is above the ground.
c) Sketch graphs of y vs. x for this motion.
d) Sketch graphs of the position, velocity and acceleration vs. time for the horizontal motion and also for
the vertical motion.
e) How would the above graphs change if, instead of being constant, the initial acceleration of the rocket
had increased uniformly from zero during the 4-sec booster phase?
The rocket boost part of the problem is a separate problem from the “free-fall” part of the problem because
of the different accelerations. You should know that the term “free-fall” means that the object is being
accelerated by gravity only, and not that it is moving downward. Use 1-D kinematics formulas to solve the
first part of the problem without breaking into components and determine the distance traveled by rocket in
the first 4 sec in the 53o direction. Then solve the free-falling second part as a 2-D projectile problem. Time
in free-fall is 17.8 sec. Total range is 1188 m.
a) 480 m
b) 21.8 s
c)
d) The first parts would reflect positive acceleration in all graphs.
The second parts would show constant velocity for the x-motion and
negative acceleration (-g) for the vertical motion.
e) Only the first part in the graphs would change. The acceleration graph would have a slopped line, the
velocity a quadratic curve and the position graph a cubic-function curve.
10. A balloon is descending with a constant speed of 12 m/s. A person in the basket hanging from the balloon
throws a stone horizontally away at a speed of 15 m/s relative to himself. The stone takes 6 s to hit the ground.
a)
b)
c)
d)
e)
Describe the motion of the stone relative to the thrower and relative to someone in the ground.
How high was the balloon above the ground when the stone was thrown?
How high is the balloon above the ground when the stone hits the ground?
Where and how fast does the stone hit the ground?
How should a person throw the stone so that it takes longer than 6 sec to hit the ground?
a) Illustrations on the right describe the motions relative to the thrower
vo
vo
and to the ground. The thrower sees the stone move away (vx=15 m/s)
and down as a projectile with no intial vertical component. The person
on the ground sees the stone with an initial velocity at an angle downwards
with respect to the horizontal (vx=15 m/s & vyo=12 m/s).
b) 252 m; c) 180 m
d) Horizontal motion looks the same to both observers: x=90m from the horizontal position of the balloon.
e) The thrower must throw the stone up-wards as well as outward.
11. A ball is thrown so that it just barely clears a wall. The wall is 3 m high and it is initially 4 m from the ball.
The initial speed of the ball is 10 m/s. Determine the two launch angles that allow the ball to clear the wall.
Explain the differences in these two cases? In which case is the ball in free-fall longer?
Combine the x formula and the y formula to generate a single formula with (tan and (tan2 in the
equation. You will need the trig identity: sec2ø =(1+ tan2ø). You can then solve for the two roots of the
quadratic and this gives the two angles: 52º and 75º.
At the lower angle the ball is on the way up when it clears the wall, and at the higher it is going down. It takes
longer in the second case.
12. It seems counterintuitive at first, but it is harder to hit a stationary target than it is to hit a target that is
falling from rest. Consider the following case: a target is initially level with the projectile and a horizontal
distance X from the projectile.
Projectile
X
Target
a) What is the angle of aim for a hit in terms of X, g, and the initial speed of the projectile vo?
b) Repeat the above question assuming that the target drops from rest at the same moment that the
projectile is fired.
c) Explain why it would be easier to hit the target in (b) .
d) If a gun fires a bullet horizontally and, at the same time, a second bullet drops from the same height,
which one hits the ground first? Justify your answer.
e) How would the answer to (d) change if air resistance were not negligible.
a) Since the projectile and target are at the same level, the range is X= [(vo2/g) sin 2. The angle of aim is the
angle under the sine function in that expression, that is: = [arcsin(Xg/ vo2)]/2.
b) In this case the angle is zero, which means that the projectile should be aimed directly at the target, as they
both drop vertically the same distance. The projectile will eventually intersect the target.
c) It’s easier to sight a target directly than try to estimate the angle that compensates for the projectile’s fall.
d) The logic of this problem suggests the two bullets would hit the ground at the same time since they travel
equal vertical distances. The fired bullet would travel longer horizontally but that doesn’t interfere with the
vertical motion.
e) Air resistance would not affect the bullets the same way because the fired bullet has the greater total
velocity and air resistance depends of the velocity squared of the object ( aair =-Cv2). Therefore the fired bullet
would take longer to hit the ground and not go as far horizontally.
You should appreciate that this is a vast simplification of the problem. We are assuming that the earth is flat,
that the fired bullet has no initial vertical component of velocity, that latitudes are not crossed, etc.. All these
things would have to be considered in the case of a real long-ranged projectile.
13. The general case of the problem above assumes that the target is initially a vertical distance H above (or
below) the projectile, as well as a horizontal distance X from it.
Target
Projectile
H
X
a) Determine the angle that ensures a hit assuming that the target drops from rest at the same instant that
the projectile is fired.
b) Does the actual speed of the projectile matter? Justify your answer.
c) Determine the angle for a hit assuming that the target remains at its original location. The solution is
similar to that of problem 11.
a) This was done in class. Write formulas for the x and y positions as functions of time for the projectile and
the target. Since the two object must be at the same place at the same time, the effect of gravity cancels out
and the angle of aim is the same as the angle of sight: tanaim= H/X
b) No, the speed doesn’t matter as long as the two objects meet above the ground.
c) This is the “stationary target” problem done in class. The angle here does depend on the speed. You need
to combine the equations: X=vo(cos)t and H= vo(cos)t –gt2/2. When you combine these equations to
eliminate the time in the air [and use the trig identity: sec2ø =(1+ tan2ø)], you get a quadratic equation in
(tan . Solving this quadratic equation gives you two possible angles. This is a messy algebraic problem, but
otherwise straightforward. Problem (11) above is a numerical version of this problem.
14. In all the previous projectile problems we assume the earth is flat when in fact we know that it is round.
a) How does the fact that the earth is round affect the time a projectile spends is flight and the range of a
projectile? Explain.
b) At what speed would a projectile fired near the surface of the earth have to move so that it would
never hit the ground and would keep orbiting (going around) the earth? Describe the acceleration of
this projectile.
c) Since the earth rotates around its axis in 24 hours, an object on the ground does the same thing. What
is the radial acceleration of an object at the equator? Is this a centripetal or a centrifugal acceleration?
d) Compare the acceleration of an object on the ground at the equator to the acceleration of an object in
orbit near the earth’s surface.
e) The moon rotates around the earth in 27.5 days and its center is 60Rearth (Rearth=6.4 x 106 )from the
center of the earth. Determine the acceleration of the moon in its orbit around the earth. Compare this
to the acceleration of an object in orbit near the surface of the earth.
a) The projectile would fall for a longer amount of time and have a longer range along the surface.
b) If the projectile is fired so fast that the curve of its trajectory matches the curvature of the earth, it will
never hit the ground and would be “in orbit around the earth”. The projectile is now executing circular
motion with the acceleration due to gravity as the centripetal acceleration. Hence, ac = g = (vt2/R). For a
near-earth orbit, the “R” is essentially the radius of the earth (6.4 x 106 m). Solving for vt = (gR)1/2 = 8 x 103
m/s (That’s about 16 x 103 mph!).
c) The radial acceleration is centripetal (toward the center). The tangential speed of an object on the equator
is (circumference of the earth) /24 hours. Determine that speed (in m/s) and then use the centripetal
acceleration formula [ac= (vt2/R)]. Answers: vt=465 m/s; ac=0.034 m/s2.
d) The acceleration in (c) is much smaller that “g” due to gravity. It is important to realize that the
acceleration due to gravity has nothing to do with the rotation of the earth.
e) This time use the circumference of the orbit of the moon (radius 60xRearth) divided by 27.5 days to find the
tangential speed of the moon in its orbit in m/s. Use the centripetal acceleration formula [ac= (vt2/R)]. This is
also much smaller than “g” at the surface. (vt=2R/T). Answers: vt=1015 m/s; ac=0.0027 m/s2.
15. The concept of "average speed" can be somewhat ambiguous when an object is turning around or moving
in an arc. Sometimes the concept refers to the magnitude of the average velocity (the overall displacement over
time). At other times the concept refers to the collection of points over which an object moves through
(regardless of direction) over time. Which concept applies should be clear from the context of the question.
s
Consider the second hand in a clock that takes 60 sec to go once around
the face of the clock. Assume the length of the second hand is 12 cm.
Starting from the 12 o'clock position the hand travels for 5 seconds.
r
The distance that the tip traverses along the arc is called “s” and the
overall displacement is called “r”. There was a similar problem in the Intro.
homework assignment.
.
a) Determine the average tangential speed (s/t) of the tip of the hand along the arc of the clock.
b) Determine the overall average speed (r/t) of the tip of the hand during this time interval.
c) Draw vinitial, vfinal, and v vectors for the tip of the hand during this time interval
d) Determine the average acceleration of the tip of the hand during this time interval. What direction is this
acceleration?
e) Compare the average acceleration of the tip to the centripetal acceleration given by the formula (ac
=v2/r). Why are they not the same?
The angle between the initial and final positions is 300= /6 radian.
a) s/t = R/t = 12(/6)/5 = 1.26 cm/s; b) r/t = 2(12)sin 15o = 1.24 cm/s.; c) Done in class.
d) a=v/t = 2(vt)sin 15o /t = 0 .650/5 = 0.13 m/s2.
e) This is less than the centripetal acceleration because the average acceleration must be taken to the limit to
get the instantaneous (centripetal) acceleration.
16. A “Ferris wheel” is a popular amusement park ride consisting of a large wheel rotating in a vertical plane
with seats along the rim where riders sit. The wheel in this problem has a 20 m diameter. During the ride, the
riders move with a uniform speed of 4 m/s around the rim. When the ride is coming to an end, the wheel
decelerates uniformly to a stop at the bottom in one full revolution.
a) Draw velocity and acceleration vectors at the points indicated in the diagram during the steady part of the
B
ride.
b) Determine the values of the velocity and acceleration at the points indicated.
C
c) Repeat the questions above for the last revolution in which the wheel comes
to rest at the bottom.
A
a) The drawings are similar to those in the circular examples in problem (1).
Beware that the direction of rotation is counterclockwise here and that
speeds are decreasing in (c).
D
E
b) During uniform motion, the tangential speed is the same at all points and has the value given in the
problem (4 m/s). The acceleration has only one value (pointing toward the center) and is given by the formula
[ac= (vt2/R)]=1.6 m/s2.
c) Because the wheel slows down the speed decreases gradually to zero in one “last” revolution that takes
about 31.4 sec. This is true because the average speed around the circumference is half the initial (2 m/s) .
Therefore there is a tangential acceleration component [at= (vt/t)= 0.127 m/s2] whose direction is
clockwise, (opposing the motion). The centripetal acceleration decreases in magnitude as the wheel slows
down from an initial value of 1.6 m/s2 to zero.
More “frame-of reference” problems to practice:
17. A man is riding a bicycle on top of a railroad flatcar that is 9 m long. The bicycle starts from rest and takes
3 seconds to traverse the length of the car. During the same time the flatcar is moving in the opposite direction
with a constant speed of 5 m/s with respect to the ground.
a)
b)
c)
d)
What is the acceleration of the rider relative to the car?…of the rider relative to the ground?
How far does the driver travel relative to the ground?
How fast is the driver moving relative to the ground when it reaches the end of the car?
Repeat the problem assuming that the flat car accelerates backwards from rest at a rate of 4 m/s2 with
respect to the ground while the bicycle accelerates as described before.
a) The acceleration with respect to the flatcar is 2 m/s2. Because the flatcar is not accelerating, the
acceleration with respect to the ground is the same. Note that the initial speed of the flatcar doesn’t affect the
acceleration of the bicycle. Acceleration depends on changes in velocity.
b) During the 3 sec, the flatcar moves backwards 15 m, so the bicycle ends up going backwards 6 m relative
to the ground (9-15= -6).
c) The final veloctiy of the bicycle relative to the flatcar is 6 m/s, but the flatcar is moving with –5 m/s
relative to the ground. So the final veloctiy of the bicycle relative to the ground is 1 m/s. Note that the average
veloctiy of the bicycle relative to the ground is -2 m/s.
Summary of answers:
vinitial
vfinal
x
a
Bicycle relative to
flatcar
0
6 m/s
9m
2 m/s2
Flatcar relative to
ground
-5 m/s
-5 m/s
-15 m
0
Bicycle relative to
ground
-5 m/s
1 m/s
-6 m
2 m/s2
d) In this case the bicycle has a different acceleration with respect to the ground. Summary of answers:
vinitial
vfinal
x
a
Bicycle relative to
flatcar
0
6 m/s
9m
2 m/s2
Flatcar relative to
ground
0
-12 m/s
-18 m
-4 m/s2
Bicycle relative to
ground
0
-6 m/s
-9 m
-2 m/s2
18. There are many versions of the “boat-and-river” problem discussed in class. Below are some variations.
Consider a boat that can travel at a top speed of 8 mph. It is trying to make it across a river that is 0.4 miles
wide and which has a current flowing with a speed of 6 mph parallel to the shore.
VR/S
VB/R
River
a) If the boat heads directly across the river, in which direction does the boat actually move relative to the
shore? Draw a velocity vector diagram illustrating the relationship between the various velocities. Be
sure to indicate the frame of reference of each velocity.
b) How long will the boat take to reach the other shore?
c) How far downstream does the boat drift? Do this in two different ways and verify your answer. Draw a
position vector diagram illustrating the relationship between distances.
d) A person in the boat drops a lifesaver into the water when the boat starts. How far is the boat from the
lifesaver when it gets to the other shore?
e) When the boat reaches the other shore it decides to return to its starting point on the other side. In
which direction must the boat point relative to the shore to achieve this? Draw a velocity vector
diagram illustrating the relationship between the various velocities. Be sure to indicate the frame of
reference of each velocity.
f) Will the return trip take more or less time? Explain.
It is very important in very important in these problems to draw separate diagrams for velocities and
distances.
a)
b) t=0.4/8 = 0.05 hour
37o
37o
Vboat/water= 8 mph
Vboat/shore= 10 mph
0.4 m
0.5 m
0.3 m
Vwater/shore= 6 mph
c) Since the distance and velocity diagrams are proportional, the drift distance is 0.4tan37o = 0.3 mile. You
can also use the velocity of the current and the time: 6x(0.05)= 0.3 mile.
d) The lifesaver also drifts with the current so it would be 0.4 m directly across from the boat.
e) The return trip involves a non-right triangle. The boat must aim upstream to compensate for the current.
You can draw a scaled drawing or use the Law of Sines or Cosines to find the value of the wanted parts of the
triangle. The angle  = 16o, and the vb/s= 2.76 mph.
f) This takes longer because the boat travels longer in the water.
Vboat/water= 8 mph
Vboat/shore

53
o
Vwater/shore= 6 mph
19. The boat above decides to try again to cross the river. This time it wants to make sure it ends up directly
across the river from where it started.
a) In which direction must the boat head relative to the shore in order to achieve this? Draw and label the
appropriate vector diagram.
b) How long will this trip take?
c) How much distance did the boat travel in the water?
d) Attempt to do the problem assuming that the speed of the boat is 6 mph and that of the current is 8 mph.
a)
b) t=0.4/5.3=0.076 hour;
49o
c) 8x(0.076)= 0.6 mile
Vboat/water= 8 mph
Vboat/shore= 5.3 mph
Vwater/shore= 6 mph
d) You cannot do the problem because you need more speed than the current to compensate for it.
20. The boat above decided to travel 0.4 miles downstream and then returned to its starting point.
a) How long will the complete trip take?
b) Compare the answer in (a) to the time it takes to go back and forth across the river as in the problem above.
c) Show that it always takes longer to travel back and forth parallel to the current than it takes to travel the
same distance back and forth perpendicular to the current.
a) The speeds add up going downstream and subtract going upstream. Total time= (0.4/14)+(0.4/2)=0.23 hr.
b) In the previous problem it takes 2(0.076)= 0.152 hr to make the round trip, so the trip parallel and antiparallel to the current takes longer.
c) You can derive general formulas for the total time in this problem and in the previous problem. If you are
good with algebra you can show that going forth and back with and against the current always takes longer
than going an equal distance back and forth across the current.
21. An airplane pilot is trying to meet his schedule by flying to his destination city 100 miles east of his
starting point in 20 minutes. However he must take into consideration that the wind is blowing directly
southeast with a speed of 20 mph.
a)
b)
c)
d)
e)
Draw a scaled velocity vector diagram for this problem.
In which direction should the pilot head his plane in the air?
What must be the speed of the plane in the air?
Would this take more or less time if there were no wind?
If he had neglected to take the wind into consideration, how far would he be off course?
a) Don’t confuse velocities values with distance values. You can only add velocities to velocities. The basic
idea is the vector relative velocity formula: vA/B + vB/C = vA/C.The desired speed is 100 mi in 1/3 hr, that is the
velocity of the plane with respect to the ground is 300 mph directed east. But the pilot has to compensate for
the wind, so the vector diagram is:
Vwind/ground= 20 mph
Vplane/wind

Vplane/ground= 300 mph
45o
b) Use scaled drawing, or vector components, or the Law of Sines or Cosines to find the value of the wanted
parts of the triangle. The angle  = 2.80o, and the vplane/wind= 286 mph.
d) No, the wind actually increases the speed of the plane relative to the ground.
e) Redraw triangle to reflect the wind adding to the 300 mph velocity. The angle of deflection is 2.58o, and the
vplane/ground= 315 mph.
45o
Vplane/wind= 300 mph

Vplane/ground
Vwind/ground= 20 mph