Homework Assignment
Solutions
5.4.1. Consider the vector ~u = 8~i − 15~j.
(a) Find a vector ~v which is in the same direction as
~u and has a magnitude of 1. Leave your answer
in exact form.
8~
~
i − 15
Answer: ~v = 17
17 j
(b) Find a vector w
~ which is in the same direction as
~u and has a magnitude of 100. Leave your answer
in exact form.
~ 1500~
Answer: w
~ = 800
17 i − 17 j
(c) Find a vector ~x which results from rotating the
vector ~u counterclockwise by an angle of π3 .
Round to two decimal places.
Answer: ~x = −0.57~i + 16.99~j
5.4.2. Two tugboats are pulling a larger boat according to the
diagram below where the filled circles are the tugboats
and the rectangle is the larger boat.
A
30◦
45◦
B
The force with which tugboat A is pulling has a magnitude of 1 200 000 N. What is the magnitude of the force
with which tugboat B must pull in order to keep the
larger boat moving parallel to the dashed line? Round
your answer to two decimal places. Hint: The large
boat travels along the resultant force vector.
Answer: 848 528.14 N
Section 5.4
(b) Find the magnitude of the gravitational force between the sun and jupiter.
Answer: 4.16 × 1023 N
(c) Find the magnitude of the gravitational force between the earth and jupiter.
Answer: 1.02 × 1018 N
5.4.4. (a) The mass of the earth is 5.972 × 1024 kg. A person is standing on the surface of the earth and her
position vector with respect to the center of the
earth is given by 5400~i + 1500~j + 3000~k. Find a
function f which describes the magnitude of the
gravitational force imparted on her by the earth;
that is if her mass is m (in kilograms) then f (m)
should be the magnitude of the gravitational force
between her and the earth. Round to two decimal
places when necessary.
Answer: f (m) = 9.86m
(b) Newton’s second law of motion says that the relationship between a force F~ and the acceleration ~a that it imparts on an object is given by
kF~ k = mk~ak where m is the mass of the object
in kilograms and the forces are in Newtons. Find
the magnitude of the acceleration due to gravity
imparted on the person in part (a) with mass m.
Answer: k~ak = 9.86
(c) Suppose the person in part (a) has a mass of
70 kg. First find the magnitude of the gravitational force between her and the earth. Next find
the force vector imparted on the center of the
earth by gravity. Round to two decimal places
when necessary. Hint: The force vector has the
same direction as the person’s position vector.
Answer: The magnitude of this force is 690.42 N
and the vector itself is
586.50~i + 162.92~j + 325.83~k
5.4.3. The following information may be helpful:
• The mass of the sun is 1.9891 × 1030 kg.
24
• The mass of the earth is 5.972 × 10
• The mass of jupiter is 1.8981 × 10
27
kg.
kg.
Suppose that the earth is at point E, the sun is at
point S, and jupiter is at point J. Also suppose that
−→
−→
→
−
SE = 149600000~i and that EJ = −538850000 i +
674200777~j (all of these units are in kilometers). Compute the values below and round your answer to two
decimal places (two decimal places in scientific notation):
(a) Find the magnitude of the gravitational force between the sun and the earth.
Answer: 3.54 × 1022 N
Last Updated: March 29, 2014
5.4.5. A small military ship is parked in a river which flows
directly north and south. A coordinate system is implemented such that ~i points directly east and has a
length of 1 mi while the vector ~j points directly north
and has a length of 1 mi. The ship’s radar detects a
helicopter as it flies near the ship. The ship then tracks
the helicopter’s movements and finds that t hours after
it passes the ship its displacement vector from the ship
is given by
(10t2 − 70t)~i + (t2 − 4)~j.
(a) How far from the ship is the helicopter after five
hours? Round to two decimal places.
Answer: 102.18 mi
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Homework Assignment
Solutions
(b) How long is it after the ship starts tracking the
helicopter that the helicopter passes the river for
the second time?
Answer: 7 h
(c) Find a vector d~ which describes the helicopter’s
displacement from its position when t = 4 to its
position when t = 8.
Answer: d~ = 200~i + 48~j
5.4.6. A triangle is displayed on a computer screen. The pixels at its corners are described by vectors p~, ~q, and ~r
where
p~ = 50~i + 100~j
~q = 50~i + 50~j
~r = 200~i + 50~j.
Recall that these position vectors describe each pixel’s
displacement from the bottom left corner of the screen
(in pixels).
Last Updated: March 29, 2014
Section 5.4
(a) It is easy to see that this triangle is a right triangle. Find the length of its hypotenuse (in pixels).
Round to the nearest pixel (no decimal places).
Answer: 158 pixels
(b) This triangle is shifted up by 100 pixels. Find
vectors p~1 , ~q1 , and ~r1 which describe the corners
of the triangle after it is shifted.
Answer: p~1 = 50~i + 200~j, ~q1 = 50~i + 150~j, and
~r1 = 200~i + 150~j
(c) After the shift in the previous part, the entire triangle is rotated about the origin by an angle of
π
~2 , ~q2 , and ~r2 which describe the
6 . Find vectors p
corners of the triangle after this rotation. Round
to the nearest pixel. Hint: Do this by rotating
each of the individual vectors.
Answer: p~2 = −57~i + 198~j, ~q2 = −32~i + 155~j,
and ~r2 = 98~i + 230~j
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