Question

an engineer is setting a speed limit on a road. she wants a car to be able to stop in 85.0ft and knows that the maximum rate at which a car can safely brake is 21.0ft/s². what initial speed (in given units) will allow the car to come a stop at the stop sign when braking with this acceleration? mph could a car traveling 45mph come to a stop before the stop sign?

Explanation:

Step1: Use the kinematic equation

The kinematic equation \(v^{2}=v_{0}^{2}+2ax\). Here, \(v = 0\) (final velocity, car stops), \(a=- 21.0\mathrm{ft/s^{2}}\) (de - acceleration), and \(x = 85.0\mathrm{ft}\).
Substitute into the equation: \(0=v_{0}^{2}+2(-21.0)(85.0)\).

Step2: Solve for \(v_{0}\)

Rearrange the equation \(v_{0}^{2}=2\times21.0\times85.0\).
\(v_{0}^{2}=3570\).
Take the square root: \(v_{0}=\sqrt{3570}\approx59.75\mathrm{ft/s}\).

Step3: Convert from \(\mathrm{ft/s}\) to \(\mathrm{mph}\)

Use the conversion factor: \(1\mathrm{ft/s}=\frac{15}{22}\mathrm{mph}\).
\(v_{0}=59.75\times\frac{15}{22}\mathrm{mph}\).
\(v_{0}=\frac{896.25}{22}\approx40.74\mathrm{mph}\).

Step4: Compare \(45\mathrm{mph}\) with the calculated speed limit

Since \(45\mathrm{mph}>40.74\mathrm{mph}\), a car traveling \(45\mathrm{mph}\) cannot come to a stop before the stop sign.

Answer:

The initial speed is approximately \(40.7\mathrm{mph}\). A car traveling \(45\mathrm{mph}\) cannot come to a stop before the stop sign.