A boat p moving due north at a speed of 5m/s.A boat Q is moving due east at a speed of 10m/s,calculate:(1)the magnitude of the velocity of P relative to Q,and(2)the bearing of the velocity of P relative to Q

Subtraqct the Q velocity comonents from the p velocity components

North component of relative motion of P relative to Q: +5 m/s
East component of relative motion of P relative to Q: -10 m/s

relative velocity magnitude = sqrt[(5)^2 + (-10)^2]
= 11.2 m/s

Bearing: tan^-1(2) = 63.4 deg west of north, or 296.6 degrees clockwise from north

To calculate the magnitude of the velocity of boat P relative to boat Q, we can use the Pythagorean theorem. The magnitude of the relative velocity can be found by taking the square root of the sum of the squares of the individual velocities.

Let's consider the velocity of boat P as Vp and the velocity of boat Q as Vq.

(1) Magnitude of the velocity of P relative to Q:
To find this, we can use the Pythagorean theorem. The magnitude of the relative velocity (Vrel) can be calculated using the formula: Vrel = sqrt(Vp^2 + Vq^2).

Vp = 5 m/s (north)
Vq = 10 m/s (east)

To use the theorem, we need to convert the velocities into a common coordinate system. Since Vp is moving north and Vq is moving east, we can convert Vp into an eastward velocity by multiplying it by the cosine of 90 degrees (since east is perpendicular to north) and convert Vq into a northward velocity by multiplying it by the cosine of 0 degrees (since north is parallel to itself). By doing this, we can compare both velocities on the same axis.

Vp_east = 5 m/s * cos(90) = 0 m/s
Vq_north = 10 m/s * cos(0) = 10 m/s

Now we can calculate the magnitude of the relative velocity:
Vrel = sqrt(Vp_east^2 + Vq_north^2)
= sqrt(0^2 + 10^2)
= sqrt(100)
= 10 m/s

Therefore, the magnitude of the velocity of boat P relative to boat Q is 10 m/s.

(2) Bearing of the velocity of P relative to Q:
The bearing is the direction of the velocity relative to a reference direction, usually measured clockwise from the north.

To determine the bearing, we can use the inverse tangent function (tan^(-1)). The formula for the bearing (B) is given by: B = atan(Vp_east / Vq_north).

Using the values we calculated earlier:
Vp_east = 0 m/s
Vq_north = 10 m/s

Bearing (B) = atan(Vp_east / Vq_north)
= atan(0 / 10)
= atan(0)
= 0 degrees

Therefore, the bearing of the velocity of boat P relative to boat Q is 0 degrees (due north).