Tammy leaves the office, drives 43 km due
north, then turns onto a second highway and continues in a direction of 30◦
north of east for 80 km.
What is her total displacement from the office?
and
At what angle is her displacement? (Consider east to be 0◦ and north 90◦.)
please help! I got 40 and 24.8 and they are both wrong ;(
90-39.9 = 50.1 Is the angle
To find Tammy's total displacement from the office, we can use vector addition.
Step 1: Convert Tammy's movement into vector components.
Tammy's movement due north of 43 km can be represented as (0, 43) km, where the x-component is 0 km and the y-component is 43 km.
Tammy's movement at 30° north of east for 80 km can be represented as (80*cos(30°), 80*sin(30°)) km.
Step 2: Add the vector components.
Adding the x-components: 0 km + (80*cos(30°)) km = 0 km + 69.28 km = 69.28 km
Adding the y-components: 43 km + (80*sin(30°)) km = 43 km + 40 km = 83 km
Step 3: Calculate the magnitude of the displacement.
The displacement is the straight-line distance between the starting point and the ending point. We can use the Pythagorean theorem to find the magnitude of the displacement.
Magnitude of the displacement = sqrt((69.28)^2 + (83)^2) = 107 km (rounded to the nearest kilometer).
Step 4: Calculate the angle of the displacement.
To find the angle, we can use inverse tangent (arctan) function.
Angle = arctan((83 km) / (69.28 km)) = 51.1° (rounded to one decimal place).
Therefore, Tammy's total displacement from the office is approximately 107 km at an angle of 51.1 degrees.
To find Tammy's total displacement from the office, we need to calculate the sum of her northward and eastward displacements.
1. Northward Displacement:
Tammy drives 43 km due north. So her northward displacement is 43 km.
2. Eastward Displacement:
Tammy then turns onto a second highway and continues in a direction 30° north of east for 80 km. To calculate her eastward displacement, we need to find the component of the 80 km distance that lies in the eastward direction, given by:
80 km * cos(30°)
Calculating this:
80 km * cos(30°) = 69.28 km (rounded to two decimal places)
So Tammy's eastward displacement is approximately 69.28 km.
3. Total Displacement:
Using the northward and eastward displacements, we can calculate the total displacement using the Pythagorean theorem. The total displacement is the magnitude of the resultant vector formed by adding the northward and eastward displacements.
Total Displacement = √[(Northward Displacement)² + (Eastward Displacement)²]
Total Displacement = √[(43 km)² + (69.28 km)²]
Calculating this:
Total Displacement = √[1849 km² + 4793.06 km²] = √(6642.06 km²) ≈ 81.5 km (rounded to one decimal place)
So Tammy's total displacement from the office is approximately 81.5 km.
4. Angle of Displacement:
To find the angle of displacement, we can use trigonometry. The angle is given by the inverse tangent of the northward displacement divided by the eastward displacement.
Angle of Displacement = arctan(Northward Displacement / Eastward Displacement)
Angle of Displacement = arctan(43 km / 69.28 km)
Calculating this:
Angle of Displacement ≈ 32.54° (rounded to two decimal places)
So the angle of Tammy's displacement, with east as 0° and north as 90°, is approximately 32.54°.
the angle that the second part makes with the first part of the path is 90+30=120º.
Using cosine law,
displacement = sqrt(43² +80²-2•43•80•cos120º) =108 km.
From the sine law
108/sin120 = 80/sinα
sinα=80•sin120/108= 0.64
α =39.9º (the angle which displacement makes with north direction)