(b) A boat heads North at 2.4 ms-1 across a river that flows East at 1.6 ms-1. (6 marks each)
(i) Calculate the resultant velocity of the boat.
(ii) The river is 30 m wide. What is the displacement from the starting point when the boat reaches the far bank?
(i) To calculate the resultant velocity of the boat, we can use vector addition. Since the boat is moving both north and east simultaneously, we can consider these two velocities as vectors.
Let's designate the northward velocity as Va (2.4 m/s) and the eastward velocity as Vb (1.6 m/s).
We can use the Pythagorean theorem to find the resultant velocity (Vr):
Vr = √(Va^2 + Vb^2)
= √((2.4 m/s)^2 + (1.6 m/s)^2)
= √(5.76 m^2/s^2 + 2.56 m^2/s^2)
= √8.32 m^2/s^2
≈ 2.88 m/s
Therefore, the resultant velocity of the boat is approximately 2.88 m/s.
(ii) To find the displacement from the starting point when the boat reaches the far bank, we need to calculate the distance traveled by the boat in the eastward direction and northward direction separately.
Distance traveled eastward:
Since the river flows east at 1.6 m/s and the 30 m width represents the distance to be crossed, we can use the formula:
Distance = Speed × Time
Time = Distance / Speed
Time = 30 m / 1.6 m/s
Time ≈ 18.75 s
Therefore, it takes approximately 18.75 seconds for the boat to cross the river.
Distance traveled northward:
We can calculate the time it takes for the boat to travel 30 m based on its northward velocity.
Time = Distance / Speed
Time = 30 m / 2.4 m/s
Time = 12.5 s
Therefore, it takes approximately 12.5 seconds for the boat to travel 30 m northward.
Since the boat is moving in both directions simultaneously, the displacement can be found by considering the time taken to travel in both directions.
Displacement = eastward distance - northward distance
Displacement = (1.6 m/s) × (18.75 s) - (2.4 m/s) × (12.5 s)
Displacement = 30 m - 30 m
Displacement = 0 m
Therefore, the displacement from the starting point when the boat reaches the far bank is 0 m.
(i) To calculate the resultant velocity of the boat, we need to find the vector sum of the boat's velocity and the river's velocity. Since both velocities are given in different directions, we can use vector addition to solve this problem.
Let's define the boat's velocity as Vb and the river's velocity as Vr.
The magnitude of Vb is given as 2.4 m/s, and its direction is North, which we can represent as 90 degrees counterclockwise from the East direction. So, Vb = 2.4 m/s at an angle of 90 degrees.
The magnitude of Vr is given as 1.6 m/s, and its direction is East, which we can represent as 0 degrees counterclockwise from itself. So, Vr = 1.6 m/s at an angle of 0 degrees.
To find the resultant velocity, we can use the Pythagorean theorem to find the magnitude, and then use trigonometry to find the direction.
Magnitude calculation:
|r| = sqrt((|Vb|^2) + (|Vr|^2))
|r| = sqrt((2.4^2) + (1.6^2))
|r| = sqrt(5.76 + 2.56)
|r| = sqrt(8.32)
|r| ≈ 2.88 m/s
Direction calculation:
tan θ = |Vr| / |Vb|
tan θ = 1.6 / 2.4
θ = arctan(0.666)
θ ≈ 33.69 degrees
Therefore, the resultant velocity of the boat is approximately 2.88 m/s at an angle of 33.69 degrees counterclockwise from the East direction.
(ii) To find the displacement from the starting point when the boat reaches the far bank, we can use the equation:
Displacement = Magnitude of the Resultant Velocity * Time
Given that the river is 30 m wide, the time taken to cross the river can be determined by dividing the width of the river by the component of the resultant velocity in the eastward direction.
Time = Width of the River / Component of Resultant Velocity in Eastward Direction
Time = 30 m / (2.88 m/s * cos(33.69 degrees))
Calculating the time:
Time = 30 m / (2.88 m/s * 0.838)
Time ≈ 13.09 s
Since the boat is only moving northward and not eastward due to the river's flow, the displacement will be the distance traveled northward. We can calculate this using:
Displacement = Magnitude of the Resultant Velocity * Time
Displacement = 2.88 m/s * 13.09 s
Displacement ≈ 37.73 m
Therefore, the displacement from the starting point when the boat reaches the far bank is approximately 37.73 meters northward.