Question

unit 2: vectors and 2d kinematics
unit 2: boat in the river problem 1
a boat can travel 6.4 m/s in still water. the boat points itself directly across a 620 meter river that has a current of 2.8 m/s.
part a
how long does it take the boat to cross the river?
96.9 s
previous answers
correct
part b
how far downstream does the boat land?
271 m
previous answers
correct
part c
what is the magnitude of the velocity of the boat relative to a person standing on the shore?
value
units

Explanation:

Step1: Identify velocities as perpendicular components

The boat's velocity across the river $v_{across}=6.4$ m/s and the river - current velocity $v_{current}=2.8$ m/s are perpendicular to each other. The velocity of the boat relative to a person on the shore $v$ can be found using the Pythagorean theorem since the two - dimensional motion components are perpendicular.

Step2: Apply Pythagorean theorem

The Pythagorean theorem for two perpendicular vectors $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ and the magnitude of their resultant $\vec{R}$ is $R = \sqrt{A^{2}+B^{2}}$. Here, $A = v_{across}$ and $B = v_{current}$, so $v=\sqrt{v_{across}^{2}+v_{current}^{2}}$.
Substitute $v_{across}=6.4$ m/s and $v_{current}=2.8$ m/s into the formula:
\[v=\sqrt{(6.4)^{2}+(2.8)^{2}}=\sqrt{40.96 + 7.84}=\sqrt{48.8}\approx6.99\] m/s

Answer:

Value: $7.0$ (rounded to one decimal place)
Units: m/s