Difficulty: Easy
Correct Answer: 4R
Explanation:
Introduction / Context:
Aerodynamic drag is a major road load on vehicles at moderate to high speeds. Understanding how drag scales with speed helps in estimating power demand, fuel economy, and acceleration performance. This question tests the classic proportionality of drag with the square of velocity.
Given Data / Assumptions:
Concept / Approach:
The aerodynamic drag force on a vehicle is modeled as Fd = 0.5 * rho * Cd * A * v^2. Holding rho, Cd, and A constant, Fd is directly proportional to v^2. Therefore, when speed doubles, the drag force becomes four times the original value.
Step-by-Step Solution:
Let Fd1 = k * v1^2 = R, where k = 0.5 * rho * Cd * A.At v2 = 2 * v1, Fd2 = k * (2 * v1)^2.Compute: Fd2 = k * 4 * v1^2 = 4 * (k * v1^2) = 4R.Hence, the new aerodynamic resistance is four times the original.
Verification / Alternative check:
Power required to overcome drag is P = Fd * v, which scales as v^3. Doubling speed would require about eight times the power, consistent with drag quadrupling and speed doubling.
Why Other Options Are Wrong:
R: Implies drag independent of speed, incorrect.2R: Would be true if drag were proportional to v (linear), not v^2.4 R^2: Dimensional nonsense; squaring force is meaningless here.R/2: Opposite of the actual trend.
Common Pitfalls:
Confusing drag force (proportional to v^2) with required power (proportional to v^3).Ignoring the assumption that coefficients and air density remain constant.
Final Answer:
4R
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