Chapter 3 Even Problem Solutions
#12.
The magnitude of the vector is:
||34î + 13ĵ|| =
p
342 + 132 = 36.4m
The angle it makes with the x-axis is:
θ = tan−1 (
#28.
The
x
component is:
Vx = 18
The
y
m
m
∗ cos(220◦ ) = −13.8
s
s
component is:
Vy = 18
#48.
13
) = 21◦
34
m
m
∗ sin(220◦ ) = −11.6
s
s
Let us make a few denitions. Let us dene
the river ows, and
x
y
to be the direction in which
to be the directly across direction. Therefore, the
~vr = .57 m
s ĵ . We will row, with respect to the
shores, with a velocity ~
v = vx î + vy ĵ . Therefore, our total velocity is:
~vT = ~v + v~r = vx î + (vy + .57 m
s )ĵ .
river ows with velocity
a.
We want our
total
velocity to be 0 in the
y
direction.
This is the
denition of rowing straight across. Therefore, we nd immediately
that:
vy = −.57
m
s
We also know that we can row with total speed of
1.3
1.3 m
s ,
therefore:
p
m q 2
= vx + vy2 = vx2 + (.57)2
s
Solving this we get that:
vx = 1.17
Because our
y
m
s
direction velocity is negative, we know we want to go
against the ow of the river, or in the
−y
the direction we want to go as the angle
θ = tan−1 (
b.
t=
1
the
x
axis:
.57
) = 26◦
1.17
We just noted that our (total) velocity in the
therefore:
direction. We can dene
below
63m
= 54s
1.17 m
s
x direction is vx = 1.17 m
s ,