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Physics (please help)

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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) -

    which has greater acceleration at t near 0?

    Alpha a = alpha/sqrt t
    which is so big near 0 that it is undefined
    beta is just an ordinaty old constant acceleration.

  • Physics (please help) -

    Try some of the Physics Tutorials:


  • What is wrong with my answer? -

    Do you see something incorrect about my reply?

  • Physics (please help) -

    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)]
    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) -

    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) -

    va = alpha 2 t^(1/2)
    then the integral of alpha 2 t^(1/2) dt
    2 alpha * (2/3) t^(3/2)
    = (4/3) alpha t^(3/2) You are right.

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