Grantham PHY220 Week 2 Assignment Problem 1

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power?

Solution:

Week 2 Problem 1

R_x=-3\:cos\:55^{\circ }=-1.720729309\:m/s

R_y=3\:sin\:55^{\circ }-2=0.4574561329\:\:m/s

The x component of the resultant is negative and the y component is positive, thus the resultant is located at the second quadrant.

R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-1.720729309\:m/s\right)^2+\left(0.4574561329\right)^2}=1.78\:m/s

\theta =tan^{-1}\left(\frac{R_y}{R_x}\right)=tan^{-1}\left(\frac{0.4574461329}{-1.720129309}\right)=-14.9^{\circ}

Therefore, the magnitude of the net velocity of the ship is 1.78 m/s, and is going 14.9 degrees North of West

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