posted by on .

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.

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.

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.

Try some of the Physics Tutorials:

http://search.yahoo.com/search?fr=mcafee&p=Physics+tutorial

Sra

• What is wrong with my answer? - ,

SraJMcGin,
Damon

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

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)

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.