Question
I have another problem that I don't know how to solve. A car travels along a straight road, heading east for 1 hour, then traveling 30 minutes on another road that leads northeast. If the car had maintained a constant speed of 40 mph, how far is it from it's starting position? I kind of know how to set it up but I don't know how to find the angle to solve it, please explain! Thank you!
Answers
Starting from (0,0)
40mi/hr * 1 hr = 40 mi east, to (40,0)
40mi/hr * 1/2 hr = 20 mi NE adds another (20/√2,20/√2)=(14.14,14.14)
So, now the car is at (54.14,14.14)
The distance
d^2 = 54.14^2 + 14.14^2 = 3131.14
d = 55.96
40mi/hr * 1 hr = 40 mi east, to (40,0)
40mi/hr * 1/2 hr = 20 mi NE adds another (20/√2,20/√2)=(14.14,14.14)
So, now the car is at (54.14,14.14)
The distance
d^2 = 54.14^2 + 14.14^2 = 3131.14
d = 55.96
You could also use the law of cosines. The angle between the two legs of the trip is 135°, so
d^2 = 40^2 + 20^2 - 2(40)(20)cos135°
= 1600+400-1131.37 = 3131.37
rounding from the approximations above accounts for the small difference.
d^2 = 40^2 + 20^2 - 2(40)(20)cos135°
= 1600+400-1131.37 = 3131.37
rounding from the approximations above accounts for the small difference.
1600+400+1131.37 = 3131.37
I'm confused with how you got 135 degrees? Please explain! Thank you for the help!
did you draw the diagram? Due east is along the x-axis. NE is at 45°.
So, the inside angle of the triangle is 180°-45° = 135°
If that kind of diagram gives you problems, you have a tough road ahead...
So, the inside angle of the triangle is 180°-45° = 135°
If that kind of diagram gives you problems, you have a tough road ahead...
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