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A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car traveling at speed s = 27.6 m/s. At the instant the speeding car passes the police car, the police car accelerates forward from rest at a constant rate of a = 2.16 m/s^2 to catch the speeding car. Assume the speeding car maintains its speed. Write an expression for the time it takes the police car to reach the speeding car. Use the variables from the problem statement in your expression.

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AMB000

Answer:

[tex]t=\frac{2s}{a}=25.56s[/tex]

Explanation:

The distance traveled by the speeding car will be [tex]d_s=st[/tex], where [tex]s=27.6m/s[/tex].

The distance traveled by the police car will be given by the formula:

[tex]d=v_{0}t+\frac{at^2}{2}[/tex], where [tex]a=2.16m/s^2[/tex] and [tex]v_0=0m/s[/tex] since it departs from rest, thus having [tex]d=\frac{at^2}{2}[/tex].

The police car will reach the speeding car when those distances are the same, or [tex]s=d[/tex], so we will have:

[tex]st=\frac{at^2}{2}[/tex]

Which means:

[tex]t=\frac{2s}{a}[/tex]

This is the expression asked, but we can use our values:

[tex]t=\frac{2s}{a}=\frac{2(27.6m/s)}{(2.16m/s^2)}=25.56s[/tex]

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