Asked by Walter
A sailboat heads north at 3 m/s for 1 hour and then tracks back to the southeast (at 45 degrees to north) at 2 m/s for 45 minutes.
a. How far has the boat sailed?
b. How far is it from its starting location?
The textbook that included the word problem is: Basic Biomechanics Sixth Edition by Susan J. Hall;
ISBN#978-0-07-337644-8
The course name: Biomechanics
a. How far has the boat sailed?
b. How far is it from its starting location?
The textbook that included the word problem is: Basic Biomechanics Sixth Edition by Susan J. Hall;
ISBN#978-0-07-337644-8
The course name: Biomechanics
Answers
Answered by
Ms. Sue
http://www.jiskha.com/display.cgi?id=1380495585
Answered by
Natalid
Please help me
Answered by
Reiny
The boat's northbound distance = 3(3600) m = 10800 m or 10.8 km
then the distance going southeast = 2(45)(60) = 5400 m or 5.4 km
Did you make a sketch?
Mine is a triangle with sides 10800 and 5400 with a 45° angle between them
a) the total distance traveled is simply the sum of these two distances = 16200 m or 16.2 km
b) Looks ideally set up for the cosine law
let the distance from end point to starting point be x m
x^2 = 10800^2 + 5400^2 - 2(5400)(10800)cos45°
...
x = √.....
= 7957.6 m
or appr 7.96 km
check my arithmetic
then the distance going southeast = 2(45)(60) = 5400 m or 5.4 km
Did you make a sketch?
Mine is a triangle with sides 10800 and 5400 with a 45° angle between them
a) the total distance traveled is simply the sum of these two distances = 16200 m or 16.2 km
b) Looks ideally set up for the cosine law
let the distance from end point to starting point be x m
x^2 = 10800^2 + 5400^2 - 2(5400)(10800)cos45°
...
x = √.....
= 7957.6 m
or appr 7.96 km
check my arithmetic
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