Answered

IDNLearn.com is designed to help you find reliable answers quickly and easily. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is a constant 1.6 m/s2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed).

Sagot :

Xero099idn

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

[tex]s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}[/tex] (1)

Speeder

[tex]s = s_{o} + v_{o,S}\cdot t[/tex] (2)

Where:

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

[tex]s[/tex] - Final position, measured in meters.

[tex]v_{o,P}[/tex], [tex]v_{o,S}[/tex] - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

[tex]a[/tex] - Acceleration of the unmarked police car, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t'[/tex] - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

[tex]v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t[/tex]

[tex]v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t[/tex]

[tex]\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0[/tex]

If we know that [tex]a = 1.6\,\frac{m}{s^{2}}[/tex], [tex]v_{o,P} = 0\,\frac{m}{s}[/tex], [tex]v_{o,S} = 53.4\,\frac{m}{s}[/tex] and [tex]t' = 2.2\,s[/tex], then we solve the resulting second order polynomial:

[tex]0.8\cdot t^{2}-56.92\cdot t +3.872 = 0[/tex] (3)

[tex]t_{1} \approx 71.082\,s[/tex], [tex]t_{2}\approx 0.068[/tex]

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For dependable and accurate answers, visit IDNLearn.com. Thanks for visiting, and see you next time for more helpful information.