Question

Abbey steers her boat in a North Easterly direction for 15 s, East for 6 s and then stops to avoid hitting a duck. If her boat travels at a constant speed of 10 m/min during this time, what is the distance and compass bearing of Abbey’s boat from the starting point when it stops? Distances are to be rounded to 2 decimal places and angle sizes to 1 decimal place.

Jethro puts his boat into the water at the same time that Abbey puts hers into the water. His boat travels at a constant speed of 12 m/min on a bearing of 140° T for 12 s and then turns and travels due South. How far is Jethro’s boat from its starting point when Abbey stops her boat for the duck? Distances are to be rounded to 2 decimal places and angle sizes to 1 decimal place.

Answer 1

First, let me say that the times given here are stupid. Who changes courses in a matter of seconds?

Abbey travels for 21 s
So, Jethro travels S for 9 s

Just draw a couple of diagrams.

Abbey travels from (0,0) on a heading of 45° for 1/4 min, at a speed of 10m/min, so she travels 2.5m at 45°, ending up at (1.77,1.77).

Then she moves E on a heading of 90° for 1/10 min, or 1m, ending up at (2.77,1.77)

She thus ends up 3.29 m away, on a heading of 57.4°

Figure Jethro's movements the same way.

Answer 2

Thanks for the answer but can this be worked out using the cosine and sine rules
Like the Cosine rule:
b^2=a^2+c^2-2ac cosB
or the Sine rule
a/sinA = b/sinB = c/sinC