PH 221
Solutions to In Class Assignment: Week 2
Winter 2017
Problem 1.
Carlos runs with a velocity of ~v = (5.0 m/s, 25◦ north of east) for 10 minutes. How far
north of his starting position does Carlos end up?
Solution
1a. Translation
Writing Carlos’s velocity as ~v , we have ~v = (5.0 m/s, 25◦ north of east). He runs
for a time ∆t = 10 minutes.
1b. Laws, Assumptions, Simplifications
We assume that his acceleration is zero and that he is running on flat ground.
2a. Physical Representation
ŷ
N
25◦
~v
E
ŷ
N
E
x̂
x̂
Velocity vector diagram
Motion diagram
2b. Mathematical Representation
∆~r
∆t
vy = v sin θ
vavg
~ =
3a. Solution
The assumption of zero acceleration implies vavg
~ = ~v . Then, we reärrange vavg
~ =
∆~
r
to get
∆t
∆~r = (∆t)vavg
~ = (10 minutes)~v .
Hence, the magnitude of displacement is
|∆r| = (10 minutes)(5.0 m/s) = 50.0 m·minutes/second = 3000 m.
31 January, 2017
PH 221
Solutions to In Class Assignment: Week 2
Winter 2017
In the chosen coördinates, the standard polar form angle for his direction of travel
is θ = 90◦ − 25◦ = 65◦ . With that, the distance north is given by
∆ry = ∆r cos θ = (3000 m) sin 65◦ ≈ 2720 m = 2.72 km.
So he ends up 2.72 km North of his original position.
4a. Evaluation
He is running at 5.0 ms, which is rather fast, for 10 minutes, so it’s reasonable
that he could have travelled 3 km overall. A direction 25◦ east of north means
that he is moving quite a bit more northward than eastward, so we expect his
northward displacement to be close to his overall displacement, but not equal to
it because he is also moving a bit easterward.
Problem 2.
The minute hand on a watch is 2.0 cm in length. What is the displacement vector of the
tip of the minute hand. . .
(a) . . . from 8:00 to 8:20 am?
Sketch of Solution
12
1
2
~ri
∆~r
◦
120
3
~rf
4
The triangle is isosceles, so the two unlabeled angles must each be 30◦ (so the angles
all sum to 180◦ ). This means the direction
of ∆~r is 30◦ East of South.
The Law of Sines tells us that the magnitude
of ∆~r satisfies |∆~r|/ sin 120◦ = |~ri |/ sin 30◦ ,
120◦
≈ 3.5 cm.
which gives |∆~r| = (2.0 cm) sin
sin 30◦
Hence, the displacement vector is
∆~r = (3.5 cm, 30◦ East of South)
(b) . . . from 8:00 to 9:00 am?
Sketch of Solution
At 8:00 am and 9:00 am, the minute hand is in the same place.
Hence, the displacement vector is ∆~r = ~0.
31 January, 2017
PH 221
Solutions to In Class Assignment: Week 2
Winter 2017
Problem 3.
A jet plane is flying horizontally with a speed of 500 m/s over a hill that slopes upward
with a 3% grade (i.e., the “rise” is 3% of the “run”). What is the component of the
plane’s velocity perpendicular to the ground?
Sketch of Solution
~v
The plane (red) is flying horizontal while the ground (green) rises (the figure is not to
scale). The two right-triangles in the above figure are similar. The 3% grade tells us that
the ratio between the two legs of either triangle √
is 0.03, which means
(by the Pythagorean
√
theorem) that the length of the hypotenuse is 12 + 0.032 = 1.0009 times that of the
longer leg. By similarity, the ratio between the magnitude of the plane’s√total velocity, ~v ,
and the component of its velocity perpendicular to the ground, v⊥ , is 1.0009/0.03. In
particular, if the plane is flying at 500 m/s, then it must have perpendicular component
v⊥ = (500 ms) √
31 January, 2017
0.03
≈ 15 m/s.
1.0009