An Australian emu is running due north in a straight line at a speed of 13.0ms and slows down to a speed of 10.6ms in 40 s. a) What is the direction of the birds

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An Australian emu is running due north in a straight line at a speed of 13.0ms and slows down to a speed of 10.6ms in 40 s.   a) What is the direction of the birds

boronganfh · Accepted Answer

(a) The emu is moving to the north with speed of υ0=13m⋅s−1 then it slows down to a speed of υ=10.6m⋅s−1 during time interval of t = 4. The difference in the speeds is negative so according to: υ→−υ0→=at→ the acceleration must be negative (the time is always positive), so its in the opposite direction of the speeds, which is to the north. And it has a magnitude of a=υ→−υ0→t =(10.6m⋅s−1)−(13m⋅s−1)4s =−0.6m⋅s−2 a=−0.6m⋅s−2 (b) Assume that the acceleration remains constant, we can get the final velocity after additional time of 2s, so t=2+4=6s, using (1), the final velocity is therefore: υ=υ0+at =(13m⋅s−1)+(−0.6m⋅s−2)(6s) =9.4m⋅s−1 the final speed is therefore: υ=9.4m⋅s−1

Karen Robbins · Accepted Answer

The acceleration of the bird is the quotient obtained when the difference of its final and initial velocities is divided by the time elapsed. a=Vf−V1 a=10.6ms−13ms4=−0.6ms2 Thus, the acceleration is equal to -0.6 m/s2 . For the velocity after 2 seconds, Vf2=Vf1+at Substituting, Vf2=10.6ms+(−0.6)×2=9.4ms

An Australian emu is running due north in a straight line at a speed o

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