Answer to Question #264669 in Physics for Von

Question

The speed of a car decreases from 40 m/s to 35 m/s at 12 meters. What is the kinetic energy of the car?

Answer 1

Let's assume uniform decceleration. Then the acceleration of the car is given as follows (see https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas):

$$a = \dfrac{v_f^2-v_i^2}{2d}$$

where $v_f = 12m/s, v_i = 40m/s$ are final and initial speeds, $d = 12m$ is the travelled distance. The speed of the car at any given moment of time $t$ is then:

$$v = v_i + at = v_i + \dfrac{v_f^2-v_i^2}{2d}t$$

In turn, the kinetic energy at any given moment of time is:

$$K = \dfrac{mv^2}{2} = \dfrac{m}{2}\left( v_i + \dfrac{v_f^2-v_i^2}{2d}t\right)^2$$

where $m$ is the mass of the car. Substituting values for $m$ and $t$ one can obtain the numeric value of kinetic enrgy at any moment of the motion.

Answer. $K = \dfrac{m}{2}\left( v_i + \dfrac{v_f^2-v_i^2}{2d}t\right)^2$.