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A jet plane is flying at a constant altitude. At time t1 = 0, it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t2 = 30.0 s, the components are vx = - 170 m/s, vy = 40 m/s. (a) Sketch the velocity vectors at t1 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.

Answer :

Answer:

a) See attachment.

b) a_avg = -8.66 i - 2.33 j

c) a = 195 degrees from + x-axis.  |a_avg| = 8.97 m/s^2

Explanation:

Given:

- Initial Velocity V_i = (90 i + 110 j)

- Final velocity V_f = (-170 i + 40 j)

- t_i = 0

- t_f = 30.0 s

Find:

(a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?

(b) the components of the average acceleration

(c) the magnitude and direction of the average acceleration.

Solution:

- The change in velocity is given by:

                            Δ V = (V_f - V_i)

                            Δ V = ( -170 i - 90 i + 40 j - 110 j)

                            Δ V = (-260 i - 70 j)

- The change in time is given as:

                            Δ t = t_f - t_i = 30 - 0 = 30 s

-The average acceleration is:

                            a_avg = Δ V / Δ t

                            a_avg = (-260 i - 70 j) / 30

                           a_avg = -8.66 i - 2.33 j

- The magnitude of acceleration is:

                            |a_avg| = sqrt ( 8.66^2 + 2.33^2)

                            |a_avg| = sqrt (80.4245)

                            |a_avg| = 8.97 m/s^2

- The direction of acceleration is given by:

                            tan (Q) = a_y / a_x

                            Q = arctan(a_y / a_x)

                            Q = arctan(2.33 / 8.66)

                            Q = 15 degrees

Hence, 180 + 15 = a.        a = 195 degrees from + x-axis.

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