How Long Does It Take for a Police Car to Catch a Speeder?

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A speeder traveling at 30.0 m/s passes a stationary police car that accelerates at 2.44 m/s². To determine when the police car overtakes the speeder, the equations of motion for both vehicles are analyzed. The police car's position is calculated using the formula x = 1/2 * a * t², while the speeder's position is x = v₀ * t. The police car catches the speeder after approximately 24.6 seconds, during which the speeder travels 738 meters. This analysis effectively illustrates the dynamics of pursuit in a straightforward physics problem.
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Homework Statement


A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44m/s2.

a. How much time passes before the speeder is taken over by the police car?
b. How far does the speeder get before bing overtaken?

Homework Equations


s=Vi+at
Vf=Vi+at
 
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Welcome to PF, sonic!
Careful with this one - there are two different kinds of motion involved. The speeder is moving at constant speed, so you need a simpler equation for the distance in that case. The equations you typed for accelerated motion are not quite right.

I suggest you make two headings, one for the speeder, one for the police car and write the appropriate equation(s) under each. Then fill in the numbers you know.

You'll need to identify what is the same for both cars and use that fact to put together a solution.

Good luck!
 
x=x0+v0t+1/2at^2

x=position
x0=initial value of position, here call it zero
v0=initial speed, 0 for the cop, 30m/s for the speeder
a=acceleration, 2.44m/s/s for the cop, 0 for the speeder


xcop=1/2*2.44m/s/s t^2

xspeeder = 30t

the cop catches the speeder when 1.22t^2 = 30t or when t=24.6s

in this time, the speeder travels 246.s x 30m/s = 738m
 
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