Question

A boat is sent across a river with a velocity of $8km{h^{ - 1}}$. If the resultant velocity of the boat is $10km{h^{ - 1}}$, the river is flowing with a velocity of
$
A. {\text{ 12}}{\text{.8km}}{{\text{h}}^{ - 1}}\
B. {\text{ 6km}}{{\text{h}}^{ - 1}}\
C. {\text{ 8km}}{{\text{h}}^{ - 1}}\
D. {\text{ 10km}}{{\text{h}}^{ - 1}}\
$

Answer 1

Hint: The velocity of flow of a river affects the boat moving through it. When a boat moves perpendicular to the direction of flow of the river, the velocities of river and the boat add up like vectors leading to the resultant change in direction of the boat.

Complete step-by-step answer:
We are given a boat which is sent across a river such that it is projected at right angles to the direction of flow of the river. The diagram represents the possible scenario where velocities of the boat and river act on each other like vectors.
${V_B}$ is the velocity of the boat and ${V_R}$ is the velocity of the river.

Since the velocity vectors of boat and river are at right angles to each other, the vectors add up according to the triangle law of vector addition which reduces to the Pythagoras theorem as the vectors are at right angles to each other.
According to the given values, we have following values for the velocities:
$
  {V_B} = 8km/h \\
  {V_{BR}} = 10km/h \\
$
We need to find the velocity of river which can be found using the Pythagoras theorem according to which
$V_{BR}^2 = V_B^2 + V_R^2$
Substituting the various values, we get
$
  {\left( {10} \right)^2} = {\left( 8 \right)^2} + V_R^2 \\
   \Rightarrow V_R^2 = 100 - 64 \\
   \Rightarrow V_R^2 = 36 \\
$
Taking the square root on both sides, we get the value of the velocity of the river.
$
  {V_R} = \pm 6km/h \\
  {\text{ or }}{V_R} = 6km/h \\
$
Hence, the correct answer is option B.

Note:
1. We ignore the negative value because magnitude of velocity cannot be negative
2. When the boat goes up the river, the net velocity of the boat decreases while if the boat goes down the river, the net velocity of the boat increases.