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Need to calculate acceleration from total distance travelled.

  • Thread starter monac
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  • #1
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Two objects move with initial velocity -7.9 m/s, final velocity 15.8 m/s, and constant accelerations.
(a) The first object has displacement 17.5 m. Find its acceleration.
Answer is 5.35 m/s^2 <-- I got this right.

(b) The second object travels a total distance of 23.0 m. Find its acceleration.
I am having trouble with this one.
So far I tried 8.14m/s^2 and 4.07 m/s^2. Webassign says it is wrong.
I don't know what I am doing wrong. :(
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi monac! welcome to pf! :wink:
… So far I tried 8.14m/s^2 and 4.07 m/s^2.
show us how you got those, and then we can see where you've gone wrong :smile:
 
  • #3
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a = change in v / change in t
t = 23m/7.9 m/s = 2.91 s <-- doesn't make sense because there are two different velocities so this approach is wrong.
a = 15.8 + 7.9 / 2.91 s
a = 8.14 m/s^2


d = v1+v2/2*t
23 = (15.8+-7.9)/2
solved for t
t = 5.82 s
plug this into vf = vi + at
15.8 + 7.9/ 5.82s
a = 4.07 m/s^2
 
  • #4
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Thank you! I never used Physics Forums before, it is my first time.
 
  • #5
tiny-tim
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hi monac! :smile:
t = 23m/7.9 m/s = 2.91 s <-- doesn't make sense because there are two different velocities so this approach is wrong.
correct! :smile: (ie, it is wrong! :redface:)

you need a constant acceleration equation with u v a and s (not u v a and t) …

if you don't know one, look up constant acceleration in the pf library or wikipedia :wink:
 
  • #6
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When I plug into v^2 = u^2 + 2as
I am getting 4.07 m/s^2 again which is the wrong answer.
And that's the only equation without a t.
 
  • #7
tiny-tim
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When I plug into v^2 = u^2 + 2as
I am getting 4.07 m/s^2 again which is the wrong answer.
And that's the only equation without a t.

hmm … yes, that's exactly the correct equation :smile:

it's also exactly the answer i get :confused:

(and just to check, it's 17.5/23.0 times the answer to the first part)

the book must be wrong :redface:

oh wait a mo …

does the question say "total distance" instead of "displacement", or is that just you? …

the total distance would be adding the negative to the positive, instead of subtracting :wink:

goodnight! :zzz:​
 
  • #8
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It's the total distance not the displacement
That's why I am having trouble with this problem
What do you mean by

"total distance would be adding the negative to the positive, instead of subtracting"?
 
  • #9
537
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try splitting part b into two parts, with zero velocity as the dividing point
 
  • #10
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so you are saying v0 = 0 m/s
vf = 15.8 m/s ?
Then what would be delta x?
15.8^2 - 0^2 / 2*23?
Do you think that's right?
 
  • #11
537
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hmm no wait, basically the 23 meters is divided into a distance traveled backwards and a distance traveled forwards. 23 meters is the total distance traveled, but what wikipedia wants is the displacement of the object.

when wikipedia uses s, it means x - x0, which is the displacement, that's not the value we're given, we're given the total distance traveled.

On wikipedia it also shows us that s = [itex]\frac{(u + v)t}{2}[/itex]

and this is the displacement for our situation, except we don't know the displacement, we only know the total distance traveled. So to get the total distance traveled, we pretend that our initial velocity is positive instead of negative.
 
  • #12
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So I understand that our initial velocity should be positive, then ... what equation do I plug it into? all equations have a t in them ... which is unknown.
 
  • #13
537
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but when s is total distance, it's not unknown, right?

total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex] basically
 
  • #14
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so
2s/u+v t
2*23/7.9+15.8 = t
1.94 s = t

vf - vi / t = a
15.8 - 7.9 / 1.94 = a
a = 4.07 m/s^2 <-- That's the wrong answer I got in the first place :(
This is disappointing.
 
  • #15
537
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well when you are no longer worried about the distinction between total distance traveled and displacement, then you no longer have the absolute value symbols there.
 
  • #16
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well when you are no longer worried about the distinction between total distance traveled and displacement, then you no longer have the absolute value symbols there.
You no longer have the absolute value symbols where? in the second part of my solution?
 
  • #17
537
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if s = [itex]\frac{(u + v)t}{2}[/itex], then total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex]

but v = u + at is still valid, it's not |v| = |u| + at, since we aren't worried about making a distinction between total distance traveled
 
  • #18
tiny-tim
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(just got up :zzz:)

As SHISHKABOB :smile: says …
hmm no wait, basically the 23 meters is divided into a distance traveled backwards and a distance traveled forwards. 23 meters is the total distance traveled, but what wikipedia wants is the displacement of the object.
… so you need to use three values of the speed instead of the usual two (in v2 = u2 + 2as) …

u = vi, v = 0, then u = 0 and v = vf,

and find separate values of |s| for the vi to 0 part, and for the 0 to vf part :wink:
 
  • #19
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if s = [itex]\frac{(u + v)t}{2}[/itex], then total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex]

but v = u + at is still valid, it's not |v| = |u| + at, since we aren't worried about making a distinction between total distance traveled
your process is still giving me 4.07 m/s^2 which is the wrong answer :(
 
  • #20
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(just got up :zzz:)

As SHISHKABOB :smile: says …


… so you need to use three values of the speed instead of the usual two (in v2 = u2 + 2as) …

u = vi, v = 0, then u = 0 and v = vf,

and find separate values of |s| for the vi to 0 part, and for the 0 to vf part :wink:
how do I get the s when I don't know the t?
 
  • #21
tiny-tim
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have you done it in two separate parts?

if so, what were your two results?
how do I get the s when I don't know the t?
there isn't a t in v2 - u2 = 2as :confused:
 
  • #22
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I am trying to find a ... what am I doing with the s ?
I am so close to giving up on this problem
 
  • #23
tiny-tim
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I am trying to find a ... what am I doing with the s ?
the total s (or, to be precise, the total |s|) is given in the question

you should get two equations, with the same acceleration (an unknown, call it a), and with different distances (call them s1 and s2, also unknowns, except that you know what |s1| + |s2| is)
 
  • #24
537
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your process is still giving me 4.07 m/s^2 which is the wrong answer :(
that's not what I'm getting, I think you need to watch your signs, be careful about which numbers are negative/positive and if you're adding/subtracting them

again it's total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex]

and then v = u +at
 

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