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 10.5 m/s in 4.20 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 1.50 s has elapsed?

Explanation:

It is given that,

Initial velocity of the bird, u = 13 m/s

Final speed of the bird, v = 10.5 m/s

Time taken, t = 4.20 s

(a) Acceleration of the bird is given by :

So, The direction of acceleration is opposite to the direction of motion.

(b) We need to find the bird’s velocity after an additional 1.50 s has elapsed i.e. t = 4.2 + 1.5 = 5.7 s. Let v' is the new final velocity.

It can be calculated using first equation of motion as :

v' = u + at

v' = 9.64 m/s

Hence, this is the required solution.

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