The distance, in feet, a moving object has traveled after t seconds is given by 2t/(4 + t). find the acceleration of the object after 5 seconds. (round your answer to three decimal places.)
1 answer:
Hi! Let me help you! a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0 a = [(2t)/(4+t)]/t <---- working equation a = {[2(5)]/9}/5 <---- cancel 5 a = 2/9 ft/s^2 <---- Answer
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Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
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