Posted by sam on Friday, May 29, 2009 at 7:50am.
we had to watch an animation which showed two balls, each traveling on a straight horizontal line at different speeds.
the graph of position vs time and velocity vs time are given.
The red ball has a linear graph and the coordinates are (0,0), (1,5), (2,10), (3,15), and (4,20)
for the blue ball:(0,0), (1,1), (2,2.5), (3,4.5), (4,7)
it is asked:
If the motion was allowed to continue the blue (lower) ball would reach the finish line (20m) at some time Δt after the red ball. What is Δt?
i though that it would take the blue ball around 9 seconds to reach the finish line,but I don't think this is right
thanks!

physics  sam, Friday, May 29, 2009 at 8:13am
here is the graph given can be seen if you search
"constant velocity versus constant acceleration" on google its the first link

physics  PC, Friday, May 29, 2009 at 9:21am
Sam, The response to
"physics  sam, Thursday, May 28, 2009 at 11:43pm"
had been posted last night, even with a suggested correction that you have made this morning.
Would you like to check back before we continue?

physics  PC, Friday, May 29, 2009 at 9:22am
physics  PC, Friday, May 29, 2009 at 12:18am
Here is a copy of the previous response.
> (0,0), (1,1), (2,2.5), (3,5), (4,7)
Could you check the data for the blue ball to see if it is not:
(0,0), (1,1), (2,2.5), **(3,4.5)**, (4,7)
If that's the case, you will find that the function of the blue ball is
y=f(x)=(x^2+3x)/4
in which case you can easily solve for
f(x)=20

physics  sam, Friday, May 29, 2009 at 9:26am
I did that, and the answer came out to be 10.57, which isn't right.
I wasn't sure if a picture of it would help at all?

physics  PC, Friday, May 29, 2009 at 12:28pm
I have uploaded a graph but I am unable to show the link as I do not have privileges to do so. Sorry.

physics  Anonymous, Friday, May 29, 2009 at 12:41pm
The answer is the solution to the equation
(x^2+3x)/4=20
Do you have difficulties finding the solution to the above equations?

physics  MathMate, Saturday, May 30, 2009 at 11:08pm
Hello Sam, here'is the link to the graph you requested:
http://i263.photobucket.com/albums/ii157/mathmate/parabola.png
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