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A drag racing car starts from rest at t = 0 and moves along a straight line with velocity given by v = bta, where b is a constant. The expression for the distance traveled by this car from its position at t=0 is:

A. b * t³
B. b * t³ / 3
C. 4672
D. 36 * t²
E. 6 * t^(3/2)

User Nefreo
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1 Answer

3 votes

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 = bt^a dt

Assuming a to equal 1 (as the other examples in the knowledge provided), we get:

bt dt = bt 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.

User JustADude
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