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.
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) To plot the path that the hiker took, we can break down the directions step by step and use a compass to determine the angles of the turns.
First, the hiker walks due east for 40 minutes. We can draw a line going straight to the right (east) for a distance of 40 minutes.
Next, the hiker changes course by going South 25 degrees West. This means that the hiker turns 25 degrees towards the west from the south direction. We can draw a line at this angle from the end point of the previous line.
After 20 minutes, the hiker changes course again at South 75 degrees East. This means the hiker turns 75 degrees towards the east from the south direction. We can draw a line at this angle from the end point of the previous line.
Finally, the hiker goes in a straight path for one hour and then heads north for 15 minutes until reaching the destination. We can draw a line going up (north) from the end point of the previous line.
B) To find the straight line distance between the starting point and the destination, we need to use the concept of vector addition. We can break down the paths into their respective components (north-south and east-west), add them up separately, and then find the magnitude of the resultant vector.
Let's assign components to each path:
Path 1 (due east for 40 minutes):
East: 40 minutes * 3.5 miles per hour = 140 miles
Path 2 (South 25 degrees West for 20 minutes):
South: 20 minutes * 3.5 miles per hour = 70 miles
West: 20 minutes * 3.5 miles per hour * sin(25 degrees) ≈ 19.1 miles
Path 3 (South 75 degrees East for one hour):
South: 60 minutes * 3.5 miles per hour = 210 miles
East: 60 minutes * 3.5 miles per hour * sin(75 degrees) ≈ 204.8 miles
Path 4 (North for 15 minutes):
North: 15 minutes * 3.5 miles per hour = 52.5 miles
Now, let's sum up the components:
North: 52.5 miles
South: 210 - 70 miles = 140 miles
East: 140 + 204.8 miles = 344.8 miles
West: 19.1 miles
To find the straight line distance, we can use the Pythagorean theorem:
Distance = √(North^2 + South^2 + East^2 + West^2)
Distance = √(52.5^2 + 140^2 + 344.8^2 + 19.1^2)
Distance ≈ 378.5 miles
Therefore, the straight line distance between the starting point and the destination is approximately 378.5 miles.