**Application Problems: Use of Addition Vector
Tools to Solve Relative Velocity**

**Vector Problems, Example One.-**

A motorboat velocity is 20 km/h in still water. If the boat must travel straight
to the nearest shore in a river whose current is 12 km/h, ¿What up stream
angle must the bow boat point at?

Before attempting to solve this problem, it
is useful to do some considerations:

- Whenever a velocity is mentioned, it is necessary to specify what is its
frame of reference to measure it. This is a case where we have relative velocities
and the tool to find the resultant or the components is the vector sum.

- It is helpful to use an identification procedure that uses two sub indexes:
the first sub index refers to the object and the second one to the frame of
reference in which that velocity is measured. In this example **V**_{BW}
is the velocity of the **B**oat relative to the **W**ater, **V**_{BS}
is the Boat velocity relative to the **S**hore and **V**_{WS}
is the Water velocity relative to the Shore. Notice **V**_{BW}
is produced by the boat motor, instead **V**_{BS} is **V**_{BW}
plus the current effect. Hence, the boat velocity relative to the shore V_{BS},
is

(A) **V**_{BS} =
**V**_{BW} + **V**_{WS}

The sentence "a motorboat velocity is 20 km/h
in still water" means V_{BW} = 20 km/h, and the sentence "a river
whose current is 12 km/h" means V_{WS} = 12 km/h. Notice **V**_{BS}
points directly straight to the opposite shore as wanted. The angle
can be obtained from the rectangular triangle in the figure:

sen = V_{WS}/V_{BW}
= 12/20 = 0.6 then
= 36.87º. The bow boat must point at an angle of 36.87º up stream in order
to cross the river directly to the other shore.

**Vector Problems, Example 2.-**

A boat velocity is 2 m/s in still water. a) If the boat points the bow straight
to the opposite shore to cross the river whose current is 1 m/s, what is the
velocity, in magnitude and direction, of the boat relative to the shore? b)
What is the boat position relative to its starting point, after 3 min?

a) The boat velocity relative to the shore
**V**_{BS}, is the sum of its velocity relative to the water **V**_{BW},
and the water velocity relative to the shore **V**_{WS} :

**V**_{BS} = **V**_{BW}
+ **V**_{WS}

As **V**_{BW} and **V**_{WS}
are the sides of a rectangular triangle,

V_{BS2} = V_{BW}^{2} + V_{WS}^{2}
or

V_{BS}^{2} = (2m/s)^{2} + (1m/s)^{2} i.e.
V_{BS} = 2.24 m/s.

Also, tan = V_{WS}/V_{BW}
= 1/2 , then =
26.57º.

b) To calculate the position after 3 min we
can use **D** = **V**t, with **D** the vector displacement, **V**
the vector velocity and t the time, an scalar. The direction of **D** is
the same as **V** and its magnitude is V·t = 2.24 m/s·3·60 s = 403.2 m.
The boat position is then at 403.2 m from the starting point and at a direction
26.57º down stream transverse to it

**Vectors, Related Sites, visit:**

**
· Vectors, Scalars
· Vectors, Scalars - Analytic Method
**

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