Question

A boat, whose speed in still water is 8.0 m/s, crosses a river with a current of 6.0 m/s. If the boat heads perpendicular to the current, what is the speed of the boat relative to an observer standing on the shore as it crosses the river?
A.) 5.3 m/s B.) 8.0 m/s C.) 6.0 m/s D.) 10.0 m/s

Answer 1

Answer:

D. 10m/s

Explanation:

Given,

speed of boat ($v_{B}$) = 8.0m/s

speed of river current ($v_{R}$) = 6.0m/s

Since the boat travels perpendicular to the current;

The resultant speed (v) of the boat, relative to an observer, will be the vector sum of the individual speeds of the boat and the current. i.e

v = $\sqrt{v_{B} ^{2} + v_{R} ^{2} }$  -------------------------(i)

Substitute the values of $v_{B}$ and $v_{R}$ into equation (i)

v =  $\sqrt{8.0^{2} + 6.0^{2} }$

v = $\sqrt{64 + 36}$

v = $\sqrt{100}$

v = 10m/s

Therefore, the speed of the boat relative to an observer standing on the shore as it crosses the river is 10m/s.

Answer 2

Answer: D

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s

Explanation:

Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.

i. 8.0m/s in the direction perpendicular to the current

ii. 6.0m/s in the direction of the current.

So, the resultant speed can be derived by using the equation;

Rs = √(Rx^2 + Ry^2)

Taking

Ry = 8.0m/s

Rx = 6.0m/s

Substituting into the equation, we have;

Rs = √(6.0^2 + 8.0^2)

Rs = √(36+64) = √100

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s