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The distance required for a vehicle to stop can be represented by \(d = rv + fv^2\), where \(d\) is the distance in metres, \(r\) is the reaction time in seconds, \(v\) is the vehicle’s velocity at the moment when the brakes are applied, in metres per second, and \(f\) is a constant that represents the amount of friction between the tires and the road (a smaller \(f\) represents more friction).

The mental state of the driver and the conditions of the road can greatly affect the values of \(r \) and \(f\). Consider the following scenario.

  • Dan drove to work under regular conditions. For this trip, \(r = 1.5\) and \(f = 0.07\).
  • It was raining heavily when Dan returned from work. The poor conditions made him pay closer attention to his driving because the road was a bit slippery. For this trip, \(r = 1.0\) and \(f = 0.09\).


  1. Write an equation to represent each trip. State any restrictions on \(v\) and \(d\).

    \(\begin{align}
     d &= 1.5v + 0.07v^2  \\
     d &= 1.0v + 0.09v^2  \end{align}\)
    \(v, d \ge 0\) for both equations

  2. Use technology to graph the system of equations. Sketch the graph of the system.


  3. What is the solution to the system of equations?

    The two curves intersect at \((0, 0)\) and \((25, 81.25)\), so these are the solutions to the system.

  4. What does/do the solution(s) represent?

    The solutions represent the velocities at which the stopping distance is the same for the two sets of conditions. At \(0 \thinspace \rm{m/s}\) it will take \(0 \thinspace \rm{m}\) to stop in both situations, and at 25 m/s it will take \(81.25 \thinspace \rm{m}\) to stop in both situations.

     For more examples on solving systems of equations graphically, see pp. 424 – 434 of Pre-Calculus 11.