Increase in speed or rate
Answer:
Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east
Explanation:
This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.
Let's take the north direction with the Y axis and the east direction as the X axis
Vy = 240 km / h airplane
Vx = 100 Km / h wind
a) See the annex
Analytical calculation of the magnitude of the speed and direction of the aircraft
V² = Vx² + Vy²
V = √ (240² + 100²)
V = 260 km/h
Airplane speed relative to the ground is 260 km/h
Tan θ = Vy / Vx
tan θ = 100/240
θ = 22.6º
Direction from north to eastb
b) What direction should the pilot have so that the resulting northbound
Vo = 240 km/h airplane
Vox = Vo cos θ
Voy = Vo sin θ
Vx = 100 km / h wind
To travel north the speeds the x axis (East) must add zero
Vx -Vox = 0
Vx = Vox = Vo cos θ
100 = 240 cos θ
θ = cos⁻¹ (100/240)
θ = 65.7º
North to West Direction
The speed in that case would be
V² = Vx² + Vy²
To go north we must find Vy
Vy² = V² - Vx²
Vy = √( 240² - 100²)
Vy = 218.2 km / h
Motion depends on what your reference point is. So a good reference point is a tree or a sign, something that is not moving. So if the zebra is walking towards the tree he/she is getting closer to the tree, He/she is in motion. But if the zebra and the tree never get closer or farther apart the zebra is not in motion. HOPE THIS HELPED!! ;D
Answer: Accelaration is 2.77 m/s*s
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Explanation:
V0=0km/h=0m/s
V1=100 km/h=27.7 m/s
t=10s=
Use equation for accelaration : a=(V1-V0)/t
a=(0m/s-27.7m/s)/10s
a=-27.7s/10s
a=2.77m/s*s