Question

A jet plane is flying at a constant altitude. At time t1 = 0, it has components of velocity vx = 88 m/s, vy = 115 m/s. At time t2 = 30s, the components are vx = -175 m/s, vy = 35 m/s.

a) Sketch the velocity vectors at t1 and t2. (figured that out on paper)

b) For this time interval calculate the components of the average acceleration, and c) the magnitude and direction of the average acceleration.

I think average acceleration (x, y) should be (-8.77, -2.67) because the x component = change in x components of velocity / change in time = (-175 - 88) / 30. And then the same respectively for y component of average acceleration.

This makes the magnitude the square root of (-8.77^2) + (-2.67^2) which = 9.17 m/s^2. Direction = inverse tangent of (-2.67 / -8.77) = 16.9, and then add 180 to put it in quadrant III.

So these answers make sense to me, but they contradict the answers in my textbook. I have had a problem with the answer key in my textbook before. I'm trying to check my work against the answer key before I submit it online, but I'm beginning to feel like the textbook answers can be misleading.

Am I correct?

Answer 1

b. V1 = 88 + 115i = 144.8m/s.[52.6o]. = Velocity at t1.
V2 = -175 + 35i = 178.5m/s[11.3o] N. of W. =178.5m/s[168.7o] CCW.

a = (V2 - V1) /(t2-t1) = (-175+35i -88-115i) /30 = -263 - 80i)/30 = -8.77- 2.67i m/s^2.

Answer 2

c. a = -8.77 - 2.67i = 9.17m/s^2[16.9o] S. of W. = 9.17m/s^2[196.9o] CCW.

My answers agree with yours.