To find Tammy's total displacement, we can break down her movements into two separate displacements: one due north and one at a 45-degree angle north of east.
For the first displacement, Tammy drives 36 km due north. Since this is a straight line, the magnitude of her displacement in the north direction is simply 36 km.
For the second displacement, Tammy travels 50 km at a 45-degree angle north of east. To find the north and east components of this displacement, we can use basic trigonometry.
The north component would be the length of the side adjacent to the 45-degree angle, and the east component would be the length of the side opposite the 45-degree angle.
Using trigonometry, we can determine that the north component (x) is given by: x = 50 km * cos(45 degrees) ≈ 50 km * 0.7071 ≈ 35.36 km
Similarly, the east component (y) is given by: y = 50 km * sin(45 degrees) ≈ 50 km * 0.7071 ≈ 35.36 km
Now we can calculate the total displacement by summing the north displacement components and the east displacement components:
Total north displacement = 36 km + 35.36 km = 71.36 km
Total east displacement = 35.36 km
To find the magnitude of the total displacement, we can use the Pythagorean theorem:
Magnitude = √(Total north displacement)^2 + (Total east displacement)^2
= √(71.36 km)^2 + (35.36 km)^2
≈ √(5099.1296 km^2 + 1250.0096 km^2)
≈ √6349.1392 km^2
≈ 79.7 km (rounded to one decimal place)
Therefore, Tammy's total displacement from the office is approximately 79.7 km.
To find the direction of the displacement, we can use trigonometry again. The direction would be the angle between the positive x-axis (east) and the displacement vector. Using inverse trigonometric functions, we find:
Angle = arctan(Total east displacement / Total north displacement)
= arctan(35.36 km / 71.36 km)
≈ arctan(0.495)
≈ 26.9 degrees (rounded to one decimal place)
So, Tammy's total displacement from the office is approximately 79.7 km in a direction of 26.9 degrees north of east.