v=14.2(12.5-s) The velocity, v, in meters per second, of an airplane s seconds after landing is given by the equation. What is the best interpretation of 14.2 as shown in the equation? Choose 1 answer: (A) The plane takes 14.2 seconds to stop. (B) The position of the plane changes 14.2 meters every second. (c) The velocity changes by 14.2 meters per second every second. (D) The velocity of the plane just as it lands is 14.2 meters per second.

Question

v=14.2(12.5-s) The velocity,  v, in meters per second, of an airplane  s seconds after landing is given by the equation. What is the best interpretation of 14.2 as shown in the equation? Choose 1 answer: (A) The plane takes 14.2 seconds to stop. (B) The position of the plane changes 14.2 meters every second. (c) The velocity changes by 14.2 meters per second every second. (D) The velocity of the plane just as it lands is 14.2 meters per second.

Bytelearn · Accepted Answer

Analyze Equation:  Analyze the equation v=14.2(12.5-s) to understand the role of 14.2. The equation describes the velocity v of an airplane in terms of seconds s after landing. Recognize Multiplication:  Recognize that 14.2 is multiplied by (12.5 - s), which suggests it's a rate of change, not a fixed time or position. Identify Unit of 14.2:  Since the unit of velocity v is meters per second and s is in seconds, the unit of 14.2 must be meters per second per second, indicating a change in velocity per second. Conclude Deceleration Rate:  Conclude that 14.2 represents the rate at which the velocity of the airplane decreases each second after landing, which matches the definition of deceleration.

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