Question

(a) In what direction would the ship have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 7.00m/s?

(b) What would its speed be relative to the Earth?

Short answer

(a) The resultant vector, velocity of ship relative to earth is7.87m/s and ship should travel at$80.6^{\circ }$north.

(b) The velocity of the ship with respect to earth is 8.04 m/s.

Step-by-step solution

Resultant velocity  

The total of an object's individual vector velocities is its resultant velocity. The scalar product of an object's mass and its acceleration vector equals the sum of the vector forces acting on it.

Given data

The ship was heading to north at7m/s relative to water.

Now the ship has to travel with some angle relative to x axis and relative to water so that, the resultant velocity towards north.

The resultant velocity will be velocity of ship relative to the earth.

Velocity of ship relative to the earth

Velocity of the ship with respect to earth

$$\begin{gathered} V_{se}=V_{sw}+V_{we} \\ \overrightarrow{R}=V_1+V_2 \end{gathered}$$

Here we need to find the x component and the y component.

$\begin{array}{l}{V}_x=V_{i}cos\theta \\ V_x=\left(7\right)\cos \theta \\ V_x=7\cos \theta \end{array}$

$$\begin{gathered} \text{Ycomponent} \\ {V}_y=V_{i}Sin\theta \\ {V}_y=\left(7\right)Sin\theta \\ {V}_y=7Sin\theta \end{gathered}$$

Resultant Vector of X component

They want to travel to straight north, so the resultant value of x component will be zero. Hence putting the value in equation

$$\begin{gathered} V_1=1.15+\left(-7\cos \theta \right) \\ 0=1.15+\left(-7\cos \theta \right) \\ \cos \theta =\frac{1.15}{7}=0.164 \\ \theta =80.6^{\circ } \end{gathered}$$

Resultant Vector of Y component

Hence the resultant vector for y component is

$\begin{array}{l}{V}_2=7\sin \theta +0.964\\ V_2=7\sin 80.6+0094\\ V_2=7.87\end{array}$

The resultant velocity is

$$\begin{gathered} \overrightarrow{R}=\sqrt{{\left(\sum _{}^x\right)}^2+{\left(\sum _{}^Y\right)}^2} \\ \overrightarrow{R}=\sqrt{{\left(0\right)}^2+{\left(7.87\right)}^2} \\ \overrightarrow{R}=7.87m/s \end{gathered}$$

The resultant vector, velocity of ship relative to earth is 7.87 m/s and ship should travel at $80.6^{\circ }$north direction.

The velocity of the ship with respect to earth is 8.04 m/s.