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.

To determine which car grabs the early lead, we need to compare their positions as a function of time. The position as a function of time can be obtained by integrating the acceleration twice.

For Team Alpha's car:
Acceleration: ax = αt^(-1/2)
Integrating once gives: vx = ∫(αt^(-1/2)) dt
= 2αt^(1/2) + C1

Integrating again gives: x = ∫(2αt^(1/2) + C1) dt
= (4/3)αt^(3/2) + C1t + C2

For Team Beta's car:
Acceleration: ax = β
Integrating once gives: vx = ∫β dt
= βt + C3

Integrating again gives: x = ∫(βt + C3) dt
= (1/2)βt^2 + C3t + C4

Now, let's compare the two positions to determine the early leader:

At t = 0, both cars start from rest, so we can set the constants of integration to zero.

For Team Alpha's car: x_alpha = (4/3)αt^(3/2)
For Team Beta's car: x_beta = (1/2)βt^2

To compare the positions, we can see that as t increases, the term t^3/2 in Team Alpha's position function will grow faster than the term t^2 in Team Beta's position function. Therefore, Team Alpha's car grabs the early lead since it starts with a higher initial position.

To find the distance x at which Team Beta's car will overtake the early leader, we set x_alpha = x_beta and solve for t:

(4/3)αt^(3/2) = (1/2)βt^2

Simplifying the equation, we get:

(8/3)α = βt^(1/2)

Solving for t, we have:

t = [(8α)/(3β)]^2

Substituting this value of t into Team Beta's position function, we can find the distance x at which Team Beta's car will overtake the early leader:

x = (1/2)β[(8α)/(3β)]^2 = (8α^2)/(9β)