A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she must stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign?

The motion of the car is defined by two stages: (i) Uniform motion, (ii) Uniform deceleration. We proceed to describe the kinematic formulas corresponding to each stage:

Uniform motion

[tex]x_{1} = x_{o} + v_{1}\cdot t[/tex] (1)

Uniform deceleration

[tex]v_{2}^{2} = v_{1}^{2} + 2\cdot a \cdot (x_{2}-x_{1})[/tex] (2)

Where:

[tex]x_{o}[/tex] - Initial position, in meters.

[tex]x_{1}[/tex] - Position when brakes are applied, in meters.

[tex]x_{2}[/tex] - Final position, in meters.

[tex]v_{1}[/tex] - Speed before brakes are applied, in meters per second.

[tex]v_{2}[/tex] - Final speed, in meters per second.

[tex]a[/tex] - Acceleration, in meters per second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{2} = 200\,m[/tex], [tex]v_{1} = 40\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.2\,s[/tex], then the deceleration needed for the car to stop at the sign is:

By (1):

[tex]x_{1} = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.2\,s)[/tex]

[tex]x_{1} = 8\,m[/tex]

And finally, by (1):

[tex]\left(0\,\frac{m}{s} \right)^{2} = \left(40\,\frac{m}{s} \right)^{2} + 2\cdot a\cdot (200\,m - 0\,m)[/tex]

[tex]a = -0.1\,\frac{m}{s^{2}}[/tex]

The car must decelerate at a rate of 0.1 meters per square second to rest at the stop sign.

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