I know how to do part A, and B..........How do you do C ? PART B: x component is

Question

I know how to do part A, and B..........How do you do C ?

PART B:

x component is -8.66m/s2

y component is -2.33m/s2

PART C

magnitude is 8.97m/s

How do i find the angle.

A jet plane is flying at a constant altitude. At time ty-0, it has components of velocity v 90 m/s, Vy = 110 m/s. At time t2 = 30.0 s, the components are v,--170 m/s, Vy = 40 m/s. (a) Sketch the velocity vectors at ty and t2 How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration

Explanation / Answer

acceleration vector

a = d(Vf - Vi)/dt

=> a = (-170 i + 40 j - 90 i - 110 j) / 30 = -8.66 i - 2.33 j m/s^2

magnitude = 8.97 m/s^2

direction tan(theta) = -2.33/-8.66 = 0.269

=> theta = 15.05 degrees from x axis

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