Final answer:
To find the distance traveled by the car, we integrate the velocity function v = bta, which results in the formula x = bt^2/2. The given options don't match this result, suggesting a possible typo or misunderstanding in the question itself.
Step-by-step explanation:
To find the expression for the distance traveled by a drag racing car with a velocity that varies as v = bta, where b is a constant and a is the car's acceleration, we need to integrate the velocity function with respect to time. Integrating v with respect to t gives us the displacement x since the car starts from rest at time t = 0.
The integral of v with respect to time t is:
∫ v dt = ∫ bta dt = b ∫ t^a dt
Assuming a to equal 1 (as the other examples in the knowledge provided), we get:
∫ bt dt = b ∫ t dt = bt^2/2 + C
Since the car starts from rest, the constant of integration C is 0. Therefore, the distance traveled is given by:
x = bt^2/2
This matches none of the options provided in the question, so it seems there may be a typo or misunderstanding in the original question or the given options. The given velocity function also lacks an exponent for t, which creates ambiguity in determining the correct answer. If a is indeed meant to represent an exponent for t, additional information is needed to determine the correct distance expression.