I couldn't find the total distance

[smashX]

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

v_f² = v_i² + 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.

 

[tiny-tim]

hi smashX!

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 … ? ​

 

[smashX]

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.

 

[tiny-tim]

the distance traveled up has to be added to the the distance traveled down to get the total distance traveled

 

[smashX]

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 :(

 

[tiny-tim]

yes, that's the idea!

so now you need to find the displacement at which the speed is zero

 

[smashX]

I think it is v_avg = (v_f + v_i) /2 = (-6+0)/2 = -3
t = (v_f - v_i)/a = 0-(-6) / 4 = 1.5s
Then s = |v_avg| * 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?

 

[tiny-tim]

hi smashX!

(oh, have a delta: ∆ … i've just realized that's what you were trying to type in your first post! )

I think it is v_avg = (v_f + v_i) /2 = (-6+0)/2 = -3
t = (v_f - v_i)/a = 0-(-6) / 4 = 1.5s
Then s = |v_avg| * 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 , but a rather strange way of doing it …

you're saying s = (v_f - v_i)/a times t

= (v_f - v_i)/a times (v_f - v_i)/2

= (v_f² - v_i²)/2a …

that's the formula you used before, so why didn't you just use it again?

 

[smashX]

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