A dog running in an open field has components of velocity v_z, = 3.4 m/s and v_y

Question

A dog running in an open field has components of velocity v_z, = 3.4 m/s and v_y = -1.8 m/s at time t_1 = 11.1 s. For the time interval from t_1 = 11.1 s to t_2 = 24.7 s, the average acceleration of the dog has magnitude 0.55 m/s^2 and direction 33.5 degree measured from the +x-axis toward the +y- axis. What is the magnitude fo the dog's velocity? Express your answer using two significant figures. What is the direction of the dog's velocity (measured from the +x-axis toward the +y- axis)? Express your answer using two significant figures.

Explanation / Answer

a = 0.55 m/s^2 ( cos33.5i + sin33.5j )

a = 0.459 i + 0.304 j m/s^2

initial velocity, u = 3.4 i - 1.8 j m/s

t = 24.7 - 11.1 = 13.6 sec

Applying, v = u + at

v = (3.4 i - 1.8 j ) + (0.459 i + 0.304 j ) (13.6)

v = 9.64 i + 2.33 j m/s

(C) magnitude = sqrt(9.64^2 + 2.33^2) = 9.92 m/s

(D) direction = tan^-1(2.33 / 9.64) = 13.6 deg fmeasured from +x -axis towards +y -axis

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