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.
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.
SraJMcGin,
Do you see something incorrect about my reply?
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.
a. To determine which car grabs the early lead, we need to compare their positions at a certain time t.
Let's integrate the acceleration functions to get the position functions for each car.
For Team Alpha's car, the acceleration is given by ax = alpha t^(-1/2). Integrating this with respect to t gives us the velocity function:
vx = ∫(alpha t^(-1/2)) dt = 2alpha t^(1/2) + C1
Integrating again, we get the position function:
x_alpha = ∫(2alpha t^(1/2) + C1) dt = (4/3)alpha t^(3/2) + C1t + C2
For Team Beta's car, the acceleration is constant at ax = Beta. Integrating this with respect to t gives us the velocity function:
vx = βt + C3
Integrating again, we get the position function:
x_beta = (1/2)βt^2 + C3t + C4
Now, to determine which car grabs the early lead, we compare their positions at t = 0. Plugging in t = 0 into both position functions, we get:
x_alpha(0) = C2
x_beta(0) = C4
Since alpha and beta are positive constants, the car with the smaller initial position, i.e., the smaller value of C2 or C4, will grab the early lead.
Therefore, the car that grabs the early lead is determined by the values of C2 and C4 in their position functions.
b. To find the distance x at which the other car will overtake the early leader, we need to set their position functions equal to each other and solve for t:
(4/3)alpha t^(3/2) + C1t + C2 = (1/2)beta t^2 + C3t + C4
Rearranging the equation:
(4/3)alpha t^(3/2) - (1/2)beta t^2 = (C3 - C1)t + (C4 - C2)
To determine the time t at which the other car overtakes the early leader, we set the coefficients of t on both sides equal:
(4/3)alpha t^(3/2) - (1/2)beta t^2 = 0
Simplifying:
(4/3)alpha t^(3/2) = (1/2)beta t^2
Dividing both sides by t^(3/2):
(4/3)alpha = (1/2)beta t^(1/2)
Solving for t:
t = (8alpha/3beta)^(2/3)
Substituting this value of t back into either position function, we can find the distance x at which the other car overtakes the early leader.