An airplane is flying with velocity (152, 0) m/s. It is accelerated in the x direction by its propeller at 8.92 m/s2 and in the negative y direction by a strong downdraft at 1.21 m/s2. What is the airplane's velocity after 3.10 seconds?
2006-11-13 15:16:06 · 2 answers · asked by xiuhcoatl 1 in Science & Mathematics ➔ Physics
its original velocity is in x direction use v=vo+at for each coordinate
then vx= 152+8.92x3.10
vy= 0 +1.21x3.10
v=sqrt(vx^2+vy^2)
2006-11-13 16:21:44 · answer #1 · answered by meg 7 · 1⤊ 0⤋
final velocity = initial velocity + acceleration*time
in the x direction:
152 + 8.92(3.1) = 179.652 m/s in the x direction
in the y direction:
0 + -1.21(3.1) = -3.751 m/s in the y direction
so in coordinate notation, the velocity of the airplane is (179.65, -3.75).
to find the magnitude, use the pythagorean therom.
179.65^2 + 3.75^2 = answer^2
magnitude of velocity = 179.68 m/s
the angle will be tan(-3.75/179.65) = -1.195
2006-11-14 00:08:43 · answer #2 · answered by Ace of Spades 2 · 0⤊ 0⤋