Find the vector components along the directions noted in parentheses.
a. a car displaced 45degrees north of east by 10.0 km (north east)
b. a duck accelerating away from a hunter at 2.0meter per second squared at an angle of 35degrees to the ground (horizontal and vertical)
I love Polka also. And Opera, and Maria CAllas.
Take the trig functions. I start the angle measurement at N, then clockwise.
a) Ncomponent=10km*cos45
Ecomponent=10km*sin45
b) Vertical component=2.0m/s^2 sin35
Horizontal component=2.0m/s^2 cos35
A. 7.1 m N
7.1 m E
B. 1.15 m vertical
1.64 m horizontal
7.7km 26 degrees
7.5km
7.1 is what I meant
To find the vector components along the given directions, we can use trigonometry.
a.
The car is displaced 45 degrees north of east, which forms a right triangle with the east and north directions.
To find the vector component along the north direction, we can use the sine function. The sine of 45 degrees is equal to the length of the opposite side divided by the length of the hypotenuse (which is the total displacement of the car, 10.0 km).
So, the north component of the car's displacement is given by:
North component = 10.0 km * sin(45 degrees) = 10.0 km * 0.7071 ≈ 7.07 km (rounded to two decimal places).
To find the vector component along the east direction, we can use the cosine function. The cosine of 45 degrees is equal to the length of the adjacent side divided by the length of the hypotenuse.
So, the east component of the car's displacement is given by:
East component = 10.0 km * cos(45 degrees) = 10.0 km * 0.7071 ≈ 7.07 km (rounded to two decimal places).
Therefore, the vector components of the car's displacement along the north and east directions are approximately 7.07 km and 7.07 km, respectively.
b.
The duck is accelerating away from the hunter at an angle of 35 degrees to the ground. We need to find the vector components along the horizontal and vertical directions.
To find the horizontal component, we can use the cosine function. The cosine of 35 degrees is equal to the length of the adjacent side (horizontal component) divided by the length of the hypotenuse (which represents the magnitude of the acceleration vector, 2.0 m/s^2).
So, the horizontal component of the acceleration is given by:
Horizontal component = 2.0 m/s^2 * cos(35 degrees) = 2.0 m/s^2 * 0.8192 ≈ 1.64 m/s^2 (rounded to two decimal places).
To find the vertical component, we can use the sine function. The sine of 35 degrees is equal to the length of the opposite side (vertical component) divided by the length of the hypotenuse.
So, the vertical component of the acceleration is given by:
Vertical component = 2.0 m/s^2 * sin(35 degrees) = 2.0 m/s^2 * 0.5736 ≈ 1.15 m/s^2 (rounded to two decimal places).
Therefore, the vector components of the duck's acceleration along the horizontal and vertical directions are approximately 1.64 m/s^2 and 1.15 m/s^2, respectively.