A plane is headed due south at 288 miles per hour.
Wind is blowing at 20 miles per hour from a direction of approximately 58 degrees.
Find the bearing for the plane.
2) A pilot at Airport A observes a plane from 35 miles away.
Airport A lies 50 miles away from Airport B.
From Airport A, the plane lies at an angle at 285deg north. Airport B lies at 65deg north.
Find the distance from the plane to Airport B.
Find the magnitude of the vector. Determine the dot product.
Do the following operation.
5)u =8i + j. v =-2i –6j. w =i – 3j.
Find the angle between vectors given to find the closest tenth of degree.
Find the amplitude and period of phase shift.
8) Find y’s amplitude = -3cos(x + p/2).
9) Find the period of the y = 2cos (2x + p/2).10) Determine the phase shift of the y = -4 +3sin (4x – 6/6)
A plane travelling at 288 mph is heading towards the south with an airspeed of 288.
From a direction of
Therefore, the bearing of a plane
A witness watches from 35 miles away, Airport A.
The distance between Airport A (and Airport B) is 50 miles.
The plane is flying at an angle 2850 from the north, which is approximately the same angle as Airport A. Airport B’s angle is approximately 650 from its north.
Angle between vector 35 mile of direction 2850 and vector 50 mile of direction 650 at airport B =650 + (900+ (2850 – 2850) = (650 + 9900 – 150) = 1400
According to the “law of cosines”, plane is approximately 32.7 miles away from Airport B.
= = = = 14.22.
The dot product for given vectors = = u.v = =u1v1 + =u2v2
u = 8i+ j
v = -2i–6j
V – (w– u).
u = 5i+ j
v = -10i + 4j
Here, t =(t1,t2) and (s1,s2)
The angle between two vectors =
The equation of the wave can be found at
For the equation, A sin = y (ot+?).
Y = A sin (ot +?). A = the sine wave’s amplitude.
The wave’s amplitude equals 3.
Period is given by T = after y = sin (ot).
The time period (T), = (as.o = 2).
For the equation, you will need to use y = A sin +?
= phase shift in the sine wave
Phase shift for a wave equation = 6.