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An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 10.9 m/s in 3.70 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 2.50 s has elapsed?

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  1. change in velocity / change in time = acceleration = (10.9 -13)/3.70
    = -.568 m/s^2
    North is positive , so the negative result means acceleration is south.
    v = Vi + a t
    = 10.9 - .568(2.50)
    = 9.48 m/s north

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  2. A jogger accelerates from rest to 3.0 m/s in 2.0s. A car accelerates from 38.0 to 41.0 m/s also in 2.0s. (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel faster than the jogger during the 2.0s? if so, how much further?

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