Posted by Abi on Friday, February 11, 2011 at 3:38pm.
Two cars are drag racing, starting from rest at t = 0, on a straight and level road. Team
Alpha's car produces acceleration
ax = alpha t^(1/2) ;
while Team Beta's car produces acceleration
ax = Beta ;
for all t > 0, where alpha and beta are positive constants with suitable units.
a. Which car grabs the early lead? Justify answer.
bIf the race goes on long enough, the other car will overtake the early leader. Find
an expression for the distance x at which this will occur, in terms of the constants alpha
and Beta.

Physics (please help)  Damon, Friday, February 11, 2011 at 3:41pm
which has greater acceleration at t near 0?
Alpha a = alpha/sqrt t
which is so big near 0 that it is undefined
whereas
beta is just an ordinaty old constant acceleration.

Physics (please help)  SraJMcGin, Friday, February 11, 2011 at 4:50pm
Try some of the Physics Tutorials:
http://search.yahoo.com/search?fr=mcafee&p=Physics+tutorial
Sra

What is wrong with my answer?  Damon, Friday, February 11, 2011 at 6:29pm
SraJMcGin,
Do you see something incorrect about my reply?
Damon

Physics (please help)  Damon, Friday, February 11, 2011 at 6:40pm
v = integral 0 to t of a dt
va = alpha 2 t^(1/2)
vb =beta t
xa = alpha (2/3) t^(3/2)
xb = beta (1/2) t^2
when xa = xb
alpha (2/3) t^(3/2) = beta (1/2) t^2
t^(1/2) = [ (4/3)(alpha/beta)]
so
t^2 = [ (4/3)(alpha/beta)]^4
and x = (1/2) beta t^2 = (1/2)beta [ (4/3)(alpha/beta)]^4

Physics (please help)  Abi, Friday, February 11, 2011 at 9:33pm
Hi Damon,
Thanks for your help. Can you explain me how you got alpha (2/3) t^(3/2) ? Because I got
alpha (4/3) t^(3/2) = beta (1/2) t^2 and then t= [(8/3)(alpha/beta)]^2 I then used my equation of beta and substituted t with my answer and I got x=(4096aplpha^4)/(162beta^3)

Physics (please help)  Damon, Friday, February 11, 2011 at 10:48pm
if
va = alpha 2 t^(1/2)
then the integral of alpha 2 t^(1/2) dt
is
2 alpha * (2/3) t^(3/2)
= (4/3) alpha t^(3/2) You are right.
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