What is the total distance traveled during the time interval?

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The discussion focuses on calculating the total distance traveled by an object under constant acceleration of 3.60 m/s², starting with an initial velocity of -6.2 m/s. The displacement for this scenario is 15.93 m, while the distance traveled differs due to the change in direction caused by the negative initial velocity. The key takeaway is that total distance accounts for the entire path traveled, including any reversal in direction, which requires a two-step calculation involving stopping distance and subsequent distance after stopping.

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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??
a-15.99m
b-15.99m
c-15.93
d-?
 
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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.
 
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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