A person walks in the following pattern: 2.8 km north, then 2.9 km west, and finally 5.5 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?
during a hockey game, a puck is given an initial speed of 10m/s. it slides 50m on the ice before it stops. what is the coeff of kin friction between the puck and ice?
S 5.5 - 2.8 = 2.7 km
W 2.9 km
magnitude = sqrt(2.7^2+2.9^2)
angle A s of w
tan A = 2.7/2.9
angle desired = 180 + A
average speed = 10/2 = 5
distance = 50
so stop time = 50/5 = 10 seconds
a = change in speed/change in time
=-10/10 = -1 m/s^2
so F = m a
mu m g = m a
mu = 1/9.81
To find the distance and angle the bird would fly in a straight line, you can use the Pythagorean theorem and trigonometry.
(a) Distance:
To find the distance, you can use the Pythagorean theorem. The person goes 2.8 km north and then 5.5 km south, so the net north-south displacement is 5.5 km - 2.8 km = 2.7 km south. The person also goes 2.9 km west.
So, the person would end up 2.7 km south and 2.9 km west from the starting point. Forming a right triangle, you can use the Pythagorean theorem:
Distance = √((2.7 km)² + (2.9 km)²)
Distance = √(7.29 km² + 8.41 km²)
Distance = √15.7 km²
Distance ≈ 3.96 km
Therefore, the bird would fly approximately 3.96 km in a straight line.
(b) Angle:
To find the angle, you can use trigonometry. Given that the person goes 2.7 km south and 2.9 km west, the angle can be found using the inverse tangent function (tan⁻¹):
Angle = tan⁻¹(opposite/adjacent)
Angle = tan⁻¹(2.7 km/2.9 km)
Angle ≈ 41.8°
Therefore, the bird would fly at an angle of approximately 41.8° counterclockwise from east.