A hiker walks 10.0 km [n], 6.8km [w], 2.4km [s] and finally 1.2km [w] in 2.5hrs.
a) What is the hiker's displacement
b) In what direction must the hiker set out, in order to return by the most direct route to the starting point?
c)if the hiker walks at a constant speed for the entire trip and returns by the most direct route, how long will the total walk take
north 10 - 2.4 = 7.6
west 6.8 + 1.2 = 8
length of distance vector= displacement = sqrt(64+57.8) = 11.0
direction from start:
tan (North/West) = 7.6 /8 = 0.95
so angle north of west = 43.5 deg which is 270 + 43.5 = 313.5 degrees clockwise from north on compass
so to head back home go 11 miles at 313.5-180 = 133.5 compass direction
total walk = 7.6 + 8 + 11
To solve this problem, we can use vector addition to find the hiker's displacement.
a) Displacement is a measure of the change in position of an object, and it includes both the magnitude (distance) and the direction. We can calculate the displacement by adding the individual displacements in each direction.
The hiker's displacement in the north direction is 10.0 km [n].
The hiker's displacement in the west direction is 6.8 km [w].
The hiker's displacement in the south direction is -2.4 km [s] (negative because it is going south).
The hiker's displacement in the west direction is -1.2 km [w] (negative because it is going west).
To find the total displacement, we can add these vectors together:
Total Displacement = 10.0 km [n] + 6.8 km [w] + (-2.4 km) [s] + (-1.2 km) [w]
To simplify the calculation, we can convert the north/south components to the same direction:
Total Displacement = 10.0 km [n] + (-2.4 km) [n] + 6.8 km [w] + (-1.2 km) [w]
Total Displacement = 7.6 km [n] + 5.6 km [w]
So, the hiker's displacement is 7.6 km [n] + 5.6 km [w].
b) To return to the starting point by the most direct route, the hiker needs to cancel out the displacement in the opposite direction. In this case, the displacement in the north direction can be canceled out by moving in the south direction, and the displacement in the west direction can be canceled out by moving in the east direction.
So, the hiker must set out in the opposite direction of the displacement, which is 7.6 km [n] + 5.6 km [w]. Therefore, the hiker should set out in the direction 7.6 km [s] + 5.6 km [e] to return to the starting point by the most direct route.
c) If the hiker walks at a constant speed and returns by the most direct route, the total walk will take the same time as the initial walk. In this case, the hiker took 2.5 hours for the entire trip. So, the total walk will also take 2.5 hours.
To answer these questions, we can use vector addition and apply simple geometry principles. Let's break down each question step-by-step:
a) To find the hiker's displacement, we need to calculate the resultant vector of all the individual displacements. We can do this by adding the vectors in the north (N) and west (W) directions separately.
The displacement in the north direction is 10.0 km [n]. This vector is purely in the north direction, so its x-axis component is 0 and its y-axis component is 10.0 km.
The displacement in the west direction is 6.8 km [w] followed by an additional 1.2 km [w]. These two vectors are in the west direction, so we can add their x-axis and y-axis components.
The x-axis component of the total westward displacement is -6.8 km - 1.2 km = -8.0 km.
The y-axis component of the total westward displacement is 0 km (since there is no displacement in the north-south direction).
Now, we can sum the x and y components of the displacements to get the resultant displacement vector.
Resultant x-axis component = -8.0 km
Resultant y-axis component = 10.0 km
Using the Pythagorean theorem, we can calculate the magnitude (distance) of the displacement:
Magnitude = √((-8.0 km)^2 + (10.0 km)^2) = √(64.0 km^2 + 100.0 km^2) = √164.0 km^2 = 12.81 km
Therefore, the hiker's displacement is approximately 12.81 km.
b) To find the direction the hiker should set out in order to return by the most direct route, we need to find the angle of the resultant displacement vector.
Angle = arctan(Resultant y-axis component / Resultant x-axis component)
Angle = arctan(10.0 km / -8.0 km)
Using a calculator, we find that the angle is approximately 124.0 degrees (rounded to the nearest whole degree). However, since the hiker needs to return by the most direct route, which is opposite to the initial displacement, we need to subtract this angle from 180 degrees.
Final angle = 180 degrees - 124.0 degrees = 56.0 degrees
Therefore, the hiker must set out at an angle of approximately 56.0 degrees (east of north) in order to return by the most direct route to the starting point.
c) To find out how long the total walk will take if the hiker walks at a constant speed for the entire trip and returns by the most direct route, we can calculate the total distance traveled and divide it by the constant speed.
Total distance traveled = 10.0 km (north) + 6.8 km (west) + 2.4 km (south) + 1.2 km (west) = 20.4 km
Let's assume the constant speed of the hiker is S km/h.
Total time = Total distance traveled / Constant speed
Total time = 20.4 km / S km/h
Therefore, the total time for the walk would be 20.4 km divided by the constant speed (S) in hours.