F10-PHYS115-UAE
Assignment 2 Solutions
1/4
27.
I
NTERPRET
We interpret this as a problem involving the addition of three displacements in two dimensions and
finding the magnitude and direction of the resultant vector. The key concepts here are displacement and average
velocity.
D
EVELOP
Using Equation 3.1, we see that in two dimensions, a displacement vector
1
r
can be written as, in
unit vector notation,
1
1
1
1
1
1
ˆ
ˆ
ˆ
ˆ
(cos
sin
)
x
y
r
r i
r
j
r
i
j
where
2
2
1
1
1
(
)
(
)
x
y
r
r
r
and
1
1
1
1
tan
(
/
).
y
x
r
r
One may write down a similar expression for
2
r
and
3
.
r
The displacement vector
1
r
is related to the velocity vector
1
v
by
1
1
.
r
v
t
We shall take a coordinate
system with
x
axis east,
y
axis north, and origin at the starting point.
E
VALUATE
(a)
The first segment of the trip which last, for
1
10 min
(1/6) h,
t
can be written, in unit-vector
notation, as
1
1
1
10
ˆ
ˆ
(40 mi/h)
h
(6.67 mi)
60
r
v
t
j
j
Similarly, we have
2
ˆ
(5.0 mi) ,
r
i
and the time spent on this segment is
2
2
2
/
(5.0 mi)/(60 mi/h)
0.083 h
5 min.
t
r v
Finally, the third segment has length
3
3
3
(30 mi/h) 6h/60
3.0 mi.
r
v
t
A unit vector in the southwest direction is
1
ˆ
ˆ
ˆ
ˆ
cos225
sin 225
(
)
2
i
j
i
j
Therefore,
3
1
ˆ
ˆ
ˆ
ˆ
(3.0 mi)
(
)
( 2.12 mi)
( 2.12 mi) .
2
r
i
j
i
j
These displacements and their sum are
shown in the figure. The total displacement is
1
2
3
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(6.67 mi)
(5.0 mi)
( 2.12 mi)
( 2.12 mi)
(2.88 mi)
(4.55 mi)
tot
r
r
r
r
v
t
j
i
i
j
i
j
(b)
Since the total travel time is
10 min
5 min
6min
21 min
(21/60) h,
t
the average velocity for the
trip is
ˆ
ˆ
(2.88 mi)
(4.55 mi)
ˆ
ˆ
(8.22 mi/h)
(13.0 mi/h)
(21/60) h
tot
r
i
j
v
i
j
t
A
SSESS
We expect both
tot
r
and
v
to be in the first quadrant since their components are all positive. Instead of
unit-vector notation,
tot
r
and
v
could be specified by their magnitudes
2
2
(2.88 mi)
(4.55 mi)
5.38 mi
tot
r
and
15.4 mi/h,
v
respectively, and common direction,
1
tan
[(4.55 mi)/(2.88 mi)]
57.7
N of E.

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