From you subject title, I will assume you have studied vectors.
let r be the resultant vector
40 min at 3.5 mph = 7/3 miles
20 min -----> 7/6 miles
60 min -----> 7/2 miles
15 min -----> 7/8 miles
r = ((7/3)cos0, (7/3)sin0) + ((7/6)cos245, (7/6)sin245) + ((7/2)cos345, ((7/2)sin345) + ((7/8)cos90, (7/8)sin90)
= (5.221 , -1.0882)
distance = |r|
= √(5.221^2 + 1.0882^2)
= 5.33 miles
or
r = v(40cos0, 40sin0) + v(20cos245, 20sin245) + v(60cos345, 60sin345) + v(15cos90, 15sin90)
r = v(89.5 , -18.655), where v is the velocity of 3.5 mph
3.5 miles/hour
= 3.5/60 miles/min
= 7/120 miles/min
distance = |r|
= (7/120)√(89.5^2 + 18.655^2))
= 5.33 miles
A hiker walks due east for 40 minutes, then changes course by going South 25 degrees West. After 20 minutes, he changes course again, this time at South 75 degrees East. He goes through this path for one hour after which he heads north reaching his final destination in 15 minutes.
A) Draw the paths that the hiker took until he reached his destination.
B) Assamuing that the hiker walked at an average speed of 3.5 miles per hour, find the straight line distance between his starting point and the destination.
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