While following the directions on a treasure
map, a pirate walks 37.8 m north, then turns
and walks 5.2 m east.
What is the magnitude of the single
straight-line displacement that the pirate
could have taken to reach the treasure?
Answer in units of m.
004 (part 2 of 2) 10.0 points
At what angle with the north would he have
to walk?
Answer in units of ◦
.
X = 5.2 m.
Y = 37.8 m.
Disp. = Sqrt(X^2 + Y^2).
Tan A = Y/X = 37.8/5.2 = 7.26923.
A = 82.2o N. of E. = 7.8o E. of N.
To find the magnitude of the single straight-line displacement that the pirate could have taken to reach the treasure, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the northward distance walked is 37.8 m and the eastward distance walked is 5.2 m. We can consider these two distances as the two sides of a right triangle. Let's call the northward distance "a" and the eastward distance "b".
Using the Pythagorean theorem, the magnitude of the single straight-line displacement (the hypotenuse) can be calculated as follows:
Magnitude = √(a^2 + b^2)
Plugging in the values, the magnitude of the single straight-line displacement is:
Magnitude = √(37.8^2 + 5.2^2) ≈ 38.1 m
Now, to find the angle at which the pirate would have to walk with the north, we can use trigonometry. The angle can be determined using the inverse tangent function (arctan).
Angle = arctan(b/a)
Plugging in the values, the angle would be:
Angle = arctan(5.2/37.8) ≈ 7.9° (rounded to one decimal place)
Therefore, the magnitude of the single straight-line displacement that the pirate could have taken to reach the treasure is approximately 38.1 meters, and the angle at which the pirate would have to walk with the north is approximately 7.9°.