Accelerated Motion: Calculating Velocity/Distance

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The discussion centers on calculating the distance and velocity of a ball rolling down an inclined plane with a constant acceleration of 2 m/s². Participants clarify that the formula d = 1/2 * a * t² gives the total distance traveled after t seconds, not the distance covered in each individual second. To find the distance traveled during a specific second, one must calculate the total distance at t seconds and subtract the total distance at (t-1) seconds. Misunderstandings arise regarding the interpretation of distances and velocities, leading to confusion about the total distance covered and the distances in each second. Ultimately, the correct approach involves using the differences in total distances to determine the distance traveled in each specific second.
  • #31
One thing: It says that the distance from zero to 2nd second is 4 meters. They also say the distance from zero to one second is 1 meter. So thinking logically I came up with 3 meters for the distance from 1 second to 2 second...Then it asks what the distance is from 2 second to 3 second(s)..then asks from 3 sec to 4 sec (length)..then from 4 sec to 5th sec...HELP!
 
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  • #32
As dx and me have been trying to tell you, to get the distance traveled in anyone second you merely have to find the total distance traveling in t seconds and the total distance traveled in t-1 seconds.

The distance traveled in the second t will be equal to the [Total distance traveled in (t) seconds]-[Total distance traveled in (t-1) seconds]

E.g take you times to be 1 second and 2 seconds.

In t seconds the total distance traveled from time t=t seconds to t=0 seconds equals
d=\frac{at^2}{2}=t^2

so:for t=1
d_1=\1^2=1

and for t=2
d_2=2^2=4

Thus for the second second, the distance traveled in was:

d_(t)-d_(t-1)=d_2-d_1= 4-1


EDIT: Dam Capslock :(
 
Last edited:
  • #33
can't see what you wrote, can you post again
 
  • #34
malty said:
As dx and me have been trying to tell you, to get the distance traveled in anyone second you merely have to find the total distance traveling in t seconds and the total distance traveled in t-1 seconds.

The distance traveled in the second t will be equal to the [Total distance traveled in (t) seconds]-[Total distance traveled in (t-1) seconds]

E.g take you times to be 1 second and 2 seconds.

In t seconds the total distance traveled from time t=t seconds to t=0 seconds equals
d=\frac{at^2}{2}=t^2

so:for t=1
d_1=1^2=1

and for t=2
d_2=2^2=4

Thus for the second second, the distance traveled was:

d_{(t)}-d_{(t-1)}=d_2-d_1= 4-1


EDIT: Dam Capslock :(

Fixed, sorry bout that :)
 
Last edited:
  • #35
So its 3 meters~!~ YA!
 
  • #36
from the second second to the third sec:
9-4 =5 metres!
 
  • #37
zachcumer said:
So its 3 meters~!~ YA!

Now, can you tell me what it would be in say the 21st second?

Let's ignore the the incline length, and please don't try to do this by adding each and everyone individually.

More importantly though, do you understand why it is the difference of the two?
 
  • #38
From the 3rd second to the fourth.

16 - 9 = 7...right?
 
  • #39
individual sec. and total sec. right?
 
  • #40
quick q:

4 plus 9 plus 16 plus 9 doesn't equal 25?
 
  • #41
zachcumer said:
quick q:

4 plus 9 plus 16 plus 9 doesn't equal 25?

Well why should it?
I've absolutely no idea what you are adding here.

1+3+5+7+9=25 is what I think you should be adding, this is true isn't it? My math may be rusty :)
 
  • #42
hey my bad ur right. Added up wrong things. Thanks so much!
 

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