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.
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.
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
To solve this problem, we will break it down into two parts: Abbey's boat and Jethro's boat.
Part 1: Abbey's Boat
First, let's determine the distance and direction Abbey's boat traveled.
1. Abbey steered her boat in a North Easterly direction for 15 seconds. Since her boat travels at a constant speed of 10 m/min, she covered a distance of (10 m/min) * (15 s / 60 s/min) = 2.5 meters.
2. Next, Abbey steered her boat East for 6 seconds, covering a distance of (10 m/min) * (6 s / 60 s/min) = 1 meter.
3. Finally, Abbey stopped her boat to avoid hitting a duck.
To find the final distance and compass bearing from the starting point, we need to use vectors. We'll represent Abbey's boat's movement as two vectors: one in the North Easterly direction (NE) and one in the East direction (E).
NE vector: magnitude = 2.5 m, angle = 45°
E vector: magnitude = 1 m, angle = 90°
Since vectors can be added using the Pythagorean theorem and vector directions can be added using trigonometry, we can find the total distance and direction from the starting point.
1. Calculate the net North component:
North component = NE magnitude * sin(NE angle) + E magnitude * sin(E angle)
North component = 2.5 * sin(45°) + 1 * sin(90°) = 1.77 m
2. Calculate the net East component:
East component = NE magnitude * cos(NE angle) + E magnitude * cos(E angle)
East component = 2.5 * cos(45°) + 1 * cos(90°) = 2.12 m
3. Use the Pythagorean theorem to find the total distance:
Total distance = sqrt(North component^2 + East component^2)
Total distance = sqrt(1.77^2 + 2.12^2) = 2.77 m (rounded to 2 decimal places)
4. Use trigonometry to find the compass bearing (angle):
Compass bearing = arctan(North component / East component)
Compass bearing = arctan(1.77 / 2.12) = 40.2° (rounded to 1 decimal place)
Therefore, Abbey's boat is approximately 2.77 meters away from the starting point, at a compass bearing of 40.2°.
Part 2: Jethro's Boat
Now let's determine how far Jethro's boat is from its starting point when Abbey stops her boat.
1. Jethro's boat travels at a constant speed of 12 m/min on a bearing of 140° T for 12 seconds.
Since Jethro's boat travels at a constant speed of 12 m/min, he covered a distance of (12 m/min) * (12 s / 60 s/min) = 2.4 meters in that direction.
2. Jethro then turns and travels due South.
Since Jethro's boat travels due South, he covers no distance in the East or West direction, only in the North direction.
Therefore, Jethro's boat is approximately 2.4 meters away from its starting point when Abbey stops her boat.