I couldn't find the total distance

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An object with a constant acceleration of 4.00 m/s² and an initial velocity of 6.00 m/s reaches a final velocity of 12.0 m/s, resulting in a displacement of 13.5 m. When the initial velocity is -6.00 m/s, the displacement remains the same, but the total distance traveled is calculated differently. The total distance includes both the upward and downward travel, leading to a total of 22.5 m after considering the time and average velocity. The calculations involve determining the time to reach the peak and then doubling the distance traveled to account for the return. The final understanding confirms the total distance traveled as 22.5 m.
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Homework Statement


An object moves with constant acceleration 4.00 m/s2 and reaches a final velocity of 12.0 m/s.
a) If its initial velocity is 6.00 m/s, find its displacement and the distance it travels
b) Same question as above instead its initial velocity is now -6.00 m/s

Homework Equations


vf2 = vi2 + 2a /\x


The Attempt at a Solution


I solve a) already and got 13.5 m as answer for both the distance and displacement. For b), the displacement is the same because plugging-in 6 or -6 doesn't change the above equation, however I don't know how to solve for the distance. I know that it will be something bigger than 13.5 m but still cannot figure out the answer a whole day. Really, really need you guy's help. Any suggestion is highly appreciated, thank you.
 
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hi smashX! :smile:

hin: imagine that gravity is 4.00 instead of 9.81 …

if you throw a ball vertically up at 6.00 m/s, then it will return to your level at the same speed, and eventually reach 12.0 m/s at the same level as if you'd thrown it downward

so to find the distance travelled, you need to find … ? :wink:
 
Total time traveled?
I got that as 4.5s and then I'm still stuck :(
Thanks tiny-tim but at the same time, sorry too, I still don't get it.
 
the distance traveled up has to be added to the the distance traveled down to get the total distance traveled :wink:
 
OK... so I draw a straight-line out and figure out that we need to find the distance from when the car is at -6m/s to 0 and then back to the position when it was at -6m/s again. That was 0 in the displacement computation... plus the 13.5 I will get the total distance right?

Somehow I get the idea from your hint, but I still don't know how to compute that distance :(
 
yes, that's the idea! :smile:

so now you need to find the displacement at which the speed is zero :wink:
 
I think it is vavg = (vf + vi) /2 = (-6+0)/2 = -3
t = (vf - vi)/a = 0-(-6) / 4 = 1.5s
Then s = |vavg| * t = 3 * 1.5 = 4.5m
The total distance will be : 2*4.5 (because it goes back and forth twice) + 13.5 = 22.5m
Is it... correct, tiny-tim?
 
hi smashX! :wink:

(oh, have a delta: ∆ … i've just realized that's what you were trying to type in your first post! :biggrin:)
smashX said:
I think it is vavg = (vf + vi) /2 = (-6+0)/2 = -3
t = (vf - vi)/a = 0-(-6) / 4 = 1.5s
Then s = |vavg| * t = 3 * 1.5 = 4.5m
The total distance will be : 2*4.5 (because it goes back and forth twice) + 13.5 = 22.5m
Is it... correct, tiny-tim?

yes :smile:, but a rather strange way of doing it …

you're saying s = (vf - vi)/a times t

= (vf - vi)/a times (vf - vi)/2

= (vf2 - vi2)/2a …

that's the formula you used before, so why didn't you just use it again? :rolleyes:
 
OK I got it now, it will be 4.5 in the end and reached the same result as above
Thank you so so much. I spent half yesterday doing this but it got to nowhere T__T . Really appreciated
 
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