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April 1, 2015

April 1, 2015

Posted by **alex** on Thursday, February 9, 2012 at 5:12pm.

westerly direction at a speed of 4 m s-1. The current of the sea is flowing at

2 m s-1 towards the north-east. Calculate how far [8 marks] and in what

direction (relative to west) [8 marks] the yacht has traveled after 10

minutes from its crossing the start line.

The answer is

(-4,0) + (1.4,1.4) = (-2.6,1.4)

= 2.96 m/s at W28.3°N

After 10 minutes (600 seconds) the distance traveled is 1772m

Can you show me how to work this out. thanks!

- maths/physics/biomechanics -
**Henry**, Saturday, February 11, 2012 at 5:15pm4m/s @ 180 Deg + 2m/s @ 45 Deg.

X = 4*cos180 + 2*cos45 = -2.586 m/s.

Y = 0 + 2*sin45 = 1.414 m/s.

tanAr=Y/X = 1.414 / -2.586 = -.054687,

Ar = -28.7 Deg. = Reference angle.

A=-28.7 + 180=151.3 Deg.,CCW= W28.7Deg

N.

V = X/cosA = -2.586 / cos151.3 = 2.95 m/s @ W28.7Deg.N.

d = 2.95m/s * 600s = 1770 m.

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