Question

A boat is heading due north across a river with aspeed of 12.0 km/h relative to the water. The water inthe river has a uniform velocity of 6.00 km/h due eastrelative to the ground. Determine the velocity of theboat relative to an observer standing on either bank.

Answer 1

Answer:

The velocity of the boat relative to an observer standing on either bank = (12î + 6j) km/h

In layman terms, the velocity of the boat relative to an observer standing on either bank is 13.42 km/h in the 26.6° North of East direction.

Explanation:

Relative velocity of body A with respect to body B, Vab, is given as Va - Vb.

where Va and Vb are velocities of A and B with respect to an external frame of reference.

That is,

Vab = Va - Vb

For this, question, let the observer be our external frame of reference (very convenient as the observer is stationary and not moving)

Let Vb = velocity of boat with respect to our external frame of reference (the observer/the ground) = ?

Va = velocity of the water of the river with respect to our external frame of reference (the observer/the ground) = 6 km/h east = (6j) km/h

Vba = velocity of the boat relative to the river = 12 km/h north = (12î) km/h

Vba = Vb - Va

12î = Vb - 6j

Vb = (12î + 6j) km/h

Magnitude = √[12² + 6²] = 13.42 km/h

Direction = tan⁻¹ (6/12) = 26.6°

Answer 2

Answer:

$v_{boat/obs}=18.0\frac{km}{h}$

Explanation:

The velocity of an object relative to a system$v_{obj/sys}$, the velocity of that system $v_{sys/obs}$ and the velocity measured by an observer at rest $v_{obj/obs}$ are related by the following equation:

$v_{obj/obs}=v_{obj/sys}+v_{sys/obs}$

In this case, the object is the boat, the system is the water and we are finding the velocity measured by an observer at rest. So, the formula becomes:

$v_{boat/obs}=v_{boat/water}+v_{water/obs}\\\\v_{boat/obs}=12.0\frac{km}{h} +6.00\frac{km}{h}=18.0\frac{km}{h}$

In words, the velocity of the boat relative to an observer standing on either bank, is 18.0km/h.