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December 19, 2014

December 19, 2014

Posted by **Abi** on Friday, February 11, 2011 at 3:38pm.

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 3:41pmwhich 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 4:50pmTry 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 6:29pmSraJMcGin,

Do you see something incorrect about my reply?

Damon

- Physics (please help) -
**Damon**, Friday, February 11, 2011 at 6:40pmv = 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 9:33pmHi 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 10:48pmif

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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