Asked by Rianna
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.
Answers
Answered by
Henry
t = 15s @ 45Deg. + 6s. @ 0 Deg.
X = 15*cos45 + 6 = 16.61 s.
Y = 15*sin45 = 10.61 s.
tanA = Y/X = 10.61 / 16.61 = 0.63877.
A = 32.57 Deg.
t =X / cos32.57 = 16.61 / cos32.57 = 19.71 s @ 32.57 Deg..
V = 10m/min = 10m/60s = 0.1667m/s.
d = V*t = 0.1667m/s * 19.71s = 3.29m
@32.57 Deg.
.
X = 15*cos45 + 6 = 16.61 s.
Y = 15*sin45 = 10.61 s.
tanA = Y/X = 10.61 / 16.61 = 0.63877.
A = 32.57 Deg.
t =X / cos32.57 = 16.61 / cos32.57 = 19.71 s @ 32.57 Deg..
V = 10m/min = 10m/60s = 0.1667m/s.
d = V*t = 0.1667m/s * 19.71s = 3.29m
@32.57 Deg.
.
Answered by
Rianna
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
Like the Cosine rule:
b^2=a^2+c^2-2ac cosB
or the Sine rule
a/sinA = b/sinB = c/sinC
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