Respuesta :
Answer:
320 km/h [E 31° S]
Explanation:
"E 42° S" means "east, 42° south", or 42° south from east.
"E 25° N" means "east, 25° north", or 25° north from east.
Taking east to be +x and north to be +y, the horizontal component of the velocity is:
vₓ = 290 cos (-42°) + 65 cos 25°
vₓ = 274 km/h
And the vertical component of the velocity is:
vᵧ = 290 sin (-42°) + 65 sin 25°
vᵧ = -167 km/h
The magnitude is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (274)² + (-167)²
v = 320 km/h
And the direction is found with trig:
θ = atan(vᵧ / vₓ)
θ = atan(-167 / 274)
θ = -31°
Therefore, the plane's velocity relative to the ground is 320 km/h [E 31° S].
The change of displacement with respect to time is defined as the velocity. The velocity of the plane relative to the ground will be 320 km/h [E 31° S]
What is velocity?
The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
Velocity at any angle is resolved to get its component of x and y-direction.
The given data in the problem is;
"E 42° S" stands for "east, 42 degrees south," or 42 degrees south of east.
"East, 25° north" or "25° north from the east" is what "E 25° N" signifies.
The following assumptions are taken for the direction purpose i.e. east as +x and north as +y
The horizontal component of the velocity is found by;
[tex]\rm v_x = 290 cos (-42^0) + 65 cos 25^0 \\\\\ \rm v_x = 274 km/h[/tex]
The velocity's vertical component is :
[tex]\rm v_ y = 290 sin (-42^0) + 65 sin 25^0 \\\\ \rm v_y = -167 km/h[/tex]
The Pythagorean theorem is used to calculate the magnitude:
[tex]\rm v=\sqrt{274^2+(-167)^2} \\\\ \rm v=320\ km/h[/tex]
The direction is found using the trigonometric formula;
[tex]\rm \theta = tan^-1( \frac{v_y}{v_x} )\\\\ \rm \theta = tan^-1( \frac{-167}{274} ) \\\\ \rm \theta =-31^0[/tex]
Hence The velocity of the plane relative to the ground will be 320 km/h [E 31° S]
To learn more about the velocity refer to the link ;
https://brainly.com/question/862972
