after t hours, the distance d between the hikers is
d^2 = (.24t)^2 + (.17t)^2 - 2(.24t)(.17t)cos109.167°
so, find t when d = 0.0568182
I get about 10 minutes.
The radios aren't much use, I guess...
Two hikers are wandering through heavy woods with walkie talkies. The walkie talkies have a range of 0.0568182 miles. From their starting point, they head off at an angle of 109.167° of each other. Hiker 1 walks 0.24 miles per hour, hiker 2 walks 0.17 miles per hour. If each continues to go straight, how long will it be before they can no longer communicate?
Can someone tell me the formula(s) to use? Please don't give the answer! Thank you!
2 answers
Given: 1000ft of range, H1 walks 1267.2ft per hour, H2 walks 897.6ft per hour, and walk from each other at approximately 109.17° angle
So to get to where b= 1000 I subtracted equally from both a(897.6) and c(1267.2) and have concluded to reach ~1001ft. Side a=415.1(H2), c= 784.7(H1)
b=√a^2+c^2-2ac*cos(B)
b= √415.1^2+784.7^2-2(415.1*784.7)*cos(109.17)
b= √172308.01+615754.09-651457.94*-.32832 →b= √788062.1+213888.4106
b= √1001950.511 →b=~ 1000.98(rounded up from 1000.97478)
For Hiker2; 897.6/x= 415.6 →60/2.15976= 27.78086
For Hiker1; 1267.2/x= 784.7 →60/1.61488= 37.15446
For Hiker2 it will take approximately 28mins, and for Hiker1 it will take approximately 37mins.
Hope this helps!
So to get to where b= 1000 I subtracted equally from both a(897.6) and c(1267.2) and have concluded to reach ~1001ft. Side a=415.1(H2), c= 784.7(H1)
b=√a^2+c^2-2ac*cos(B)
b= √415.1^2+784.7^2-2(415.1*784.7)*cos(109.17)
b= √172308.01+615754.09-651457.94*-.32832 →b= √788062.1+213888.4106
b= √1001950.511 →b=~ 1000.98(rounded up from 1000.97478)
For Hiker2; 897.6/x= 415.6 →60/2.15976= 27.78086
For Hiker1; 1267.2/x= 784.7 →60/1.61488= 37.15446
For Hiker2 it will take approximately 28mins, and for Hiker1 it will take approximately 37mins.
Hope this helps!
0 answers