Example

Question:

Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It depends on the initial velocity (\(v_0\)) and the braking capacity, or deceleration, \(-a\). Derive an expression for stopping distance of a vehicle in terms of \(v_0\) and \(a\).

Solution:

Let the distance travelled by the vehicle before it stops be \(d_s\). Using the equation of motion \[ v^2 = v_0^2 + 2a x \] and noting that for stopping, \(v = 0\), \[ 0 = v_0^2 + 2a d_s \implies d_s = -\frac{v_0^2}{2a} \] (where \(a\) is negative because it is deceleration).
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling \(v_0\) increases the stopping distance by a factor of 4 (for the same deceleration).

For a particular car, stopping distances of 10 m, 20 m, 34 m, and 50 m were found for velocities of 11, 15, 20, and 25 m/s respectively, which are nearly consistent with this result.

Stopping distance is important in setting speed limits, for example in school zones.