A car travelling at 30.0 m/s brakes on dry concrete at 7.00 m/s^2 until it stops. What distance does it cover while braking?

1. Identify the knowns: the initial velocity is 30.0 m/s, the final velocity is 0 and the acceleration is -7.00 m/s^2. 2. The acceleration is negative because it opposes the velocity. 3. Choose the equation that links velocity, acceleration and displacement, since the stopping time is unknown. 4. Rearranged, the displacement equals the difference of the squared velocities divided by twice the acceleration. 5. Substitute the values: (0 - (30.0 m/s)^2) / (2 x (-7.00 m/s^2)). 6. This gives (-900 m^2/s^2) / (-14.0 m/s^2) = 64.3 m. 7. Dividing 30.0 by 7.00 gives 4.29, which is a time in seconds and not a distance. _Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.5 Motion Equations for Constant Acceleration in One Dimension_