$v=14.2(12.5-s)$ $\newline$ 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? $\newline$ Choose $1$ answer: $\newline$ (A) The plane takes $14$ . $2$ seconds to stop. $\newline$ (B) The position of the plane changes $14$ . $2$ meters every second. $\newline$ (c) The velocity changes by $14$ . $2$ meters per second every second. $\newline$ (D) The velocity of the plane just as it lands is $14$ . $2$ meters per second.

Full solution

v=14.2(12.5-s)

\newline

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?

\newline

Choose

1

answer:

\newline

(A) The plane takes

14

2

seconds to stop.

\newline

(B) The position of the plane changes

14

2

meters every second.

\newline

14

2

meters per second every second.

\newline

(D) The velocity of the plane just as it lands is

14

2

meters per second.

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.