A boy wishes to row across a river in the shortest possible time. He can row at 2m/s in still water and the river is flowing at 1m/s. At what angle should he point th efront of his boat?

The answer is 90 degrees but I don't understand how to solve the problem using the velocity components of the boats velocity.

Can someone explain?

The boy should point the front of his boat at a 90 degree angle to the direction of the river flow. This will allow him to take advantage of the river's current, while still rowing in a straight line.

To solve this problem using velocity components, you need to consider the velocity of the boat relative to the river. The boat's velocity relative to the river is the vector sum of the boat's velocity in still water (2m/s) and the river's velocity (1m/s). This vector sum has a magnitude of 3m/s and an angle of 90 degrees relative to the river's flow. Therefore, the boy should point the front of his boat at a 90 degree angle to the direction of the river flow.

Ahoy there! Let me navigate you through this river-crossing dilemma with a touch of humor.

To find the angle at which the boy should aim his boat, we need to consider the velocity components of the boat. Now, imagine you're on a boat, and to find the quickest route across the river, you must defy the flow of the current while maximizing your own speed.

So, if the river is flowing at 1m/s and our champion rower can paddle at 2m/s in still water, we see that the river current can tangentially hinder his progress. But fear not, my friend, for our sailor can overcome this obstacle!

To ensure the shortest time, the boy must aim his boat perpendicular to the flow of the river. In other words, imagine trying to reach the other side by rowing directly across the river, forming a right angle between your heading and the water's flow.

By doing this, the flowing river current only affects the boat's sideways motion, which has no impact on the time taken to reach the other side. Meanwhile, the boat's forward velocity of 2m/s counteracts the river's current efficiently, leading to a straight and brisk crossing!

So, set your course at a bold 90 degrees to the river's flow, and our little rower will conquer the waters in the shortest possible time!

Stay afloat with laughter, my friend!

To solve this problem using the velocity components, you need to understand the concept of vector addition. The boy's boat will have two velocity components: one in the direction of the river's flow and one perpendicular to it.

Let's assume the width of the river is d. The velocity of the boat relative to the water is 2m/s in still water, and the river's flow is at 1m/s. Therefore, the velocity of the boat's component in the direction of the river's flow is 2m/s - 1m/s = 1m/s (since the boat is rowing against the river's flow).

The velocity component perpendicular to the river's flow is the speed of the boat relative to the water in the perpendicular direction, which is still 2m/s.

To find the shortest possible time, the boat should row perpendicular to the river's flow. This means the angle between the front of the boat and the direction of the river's flow should be 90 degrees.

When the boat rows perpendicular to the river's flow, the component of the boat's velocity in the direction of the river cancels out the river's flow velocity. Therefore, the only velocity that contributes to the crossing of the river is the perpendicular component, which is 2m/s.

By rowing perpendicular to the river, the boy minimizes the time it takes to cross the river.

To find the angle at which the boy should point the front of his boat, we can utilize the concept of vector addition. The boy's velocity has two components: the velocity of the boat in still water (2 m/s) and the velocity of the river current (1 m/s).

We can represent these velocities as vectors: v_boat = 2 m/s and v_river = 1 m/s. The resultant vector, which represents the boat's velocity relative to the ground, can be found by adding these two vectors together.

The magnitude of the resultant vector can be calculated using the Pythagorean theorem. Since the magnitude of v_boat = 2 m/s and v_river = 1 m/s, we have:

Resultant magnitude (v_resultant)^2 = (v_boat)^2 + (v_river)^2
= 2^2 + 1^2
= 4 + 1
= 5

Thus, the magnitude of the resultant vector is √5 m/s.

To find the angle at which the boy should point the front of his boat, we need to determine the direction of the resultant vector relative to the horizon. This can be achieved by considering the tangent of the angle.

Tangent (θ) = v_river / v_boat
= 1 / 2
= 0.5

To find the value of θ, we take the inverse tangent (arctan) of both sides:

θ = arctan(0.5)
≈ 26.57 degrees

However, keep in mind that this is the angle of the resultant vector relative to the horizontal direction, which is not the desired angle.

To determine the angle at which the boy should point the front of his boat, we need to find the complementary angle. Since the resultant vector and the desired direction will form a right angle (90 degrees), the angle the boy should point his boat is the complementary angle of 26.57 degrees:

90 degrees - 26.57 degrees = 63.43 degrees

Therefore, the boy should point the front of his boat at approximately 63.43 degrees to row across the river in the shortest possible time.