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A police car at rest, passed by a speeder traveling at a constant 120 km/h, takes off in hot pursuit. The police officer catches up to the speeder in 750 m, while maintaining a constant acceleration. Calculate (a) how long it took the police car to overtake the speeder, (b) the required police acceleration, and (c) the velocity of the police car at the moment it reaches the speeder.

Respuesta :

lcmendozaf

Answer:

a) t = 22.5 seconds

b) [tex]a = 2.96 m/s^{2}[/tex]

c) [tex]V_{p} = 66.6 m/s[/tex]  in the same direction of the car.

Explanation:

First of all, let's convert the speed of the car to m/s:

[tex]V_{c}=120km/h*1000m/km *1h/3600s =33.3m/s[/tex]

Now, since the police officer catches the car, we know that their position is the same 750m, so:

[tex]X_{p}=X_{c}=V_{c}*t[/tex]

[tex]750=33.3*t[/tex]    Solving for t, we get:

t=22.5s   Solved part a)

For the acceleration:

[tex]X_{p}=V_{op}*t+\frac{a*t^{2}}{2}[/tex]  Replacing values and solving for a:

[tex]a=2.96m/s^{2}[/tex]   Solved part b)

For the velocity:

[tex]V_{p}=V_{op}+a*t=0+2.96*22.5=66.6m/s[/tex]  Solved part c)

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