|Jun19-12, 09:33 PM||#1|
total distance traveled
An object moves with constant acceleration 3.60 m/s2 and over a time interval reaches a final velocity of 12.4 m/s.
(a) If its initial velocity is 6.2 m/s, what is its displacement during the time interval?
(b) What is the distance it travels during this interval?
(c) If its initial velocity is −6.2 m/s, what is its displacement during the time interval?
(d) What is the total distance it travels during the interval in part (c)?
I have already solve parts a, b, and c. For some reason d is tricking me though. is total distance different than distance traveled in part b??
|Jun19-12, 09:58 PM||#2|
Yes, the distance traveled in d is different than the distance traveled in b, the object travels a certain distance backwards in d since the initial velocity is negative.
|Jun20-12, 05:31 AM||#3|
The standard kinematic equations always give displacement. When initial velocity is opposite to acceleration, particle will stop some time in future and then reverse its direction. If the time considered is less than time of stopping then distance will be same as the disp, but when time involved is greater than the time of stopping then you should calculate in two steps:
Displacement while stopping and displacement from stopping to the time considered. Add them and get the answer.
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