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Resources Problem 2.38 An unmarked police car traveling a constant 85 km/h is pa

ID: 1524390 • Letter: R

Question

Resources Problem 2.38 An unmarked police car traveling a constant 85 km/h is passed by a speeder traveling 137 km/h esources previous l 21 of 22 I next Part A Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator: if the police car's acceleration is 2.30 m/s how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)? Express your answer with the appropriate units. 4.55 Submit My Answers Give Up incorrect; Try Again; 2 attempts remaining Provide Feedback Continue kangaroo ju

Explanation / Answer

We convert the units:

(85km/h)/(3.6ks/h) = 23.61 m/s.

(137km/h)/(3.6ks/h) = 38.06 m/s.

We use a coordinate system with the origin where the motorist passes the police officer.

The location of the speeding motorist is given by

  xm = xo + vmt = 0 + (38.06 m/s)t.

The location of the police officer is given by

xp = xo + vop(1.00s) + vop(t - 1.00s) + 1/2ap(t< - 1.00s)2= 0 + (23.61 m/s)t + 1/2(2.3m/s2)(t - 1.00s)2.

e need to consider how much time passes while the distance decreases between the two cars. Since we do not know

the final speed of the police car when it overtakes the speeder, we need to consider the distance driven as it relates to

time. Consider the following equivalent statements:

distance driven by speeder = distance driven by police car

distance while driving faster than police + distance while driving slower than police = distance while driving slower

than speeder + distance while driving faster than speeder

distance while driving faster than police at constant speed+ distance while driving slower than police at constant

speed= distance while driving slower than speeder at constant speed+distance while driving slower than speeder with

acceleration+ distance while driving faster than speeder with acceleration

xm = xp;

(38.06m/s)t = (23.61m/s)t + 1/2(2.30m/s2)(t - 1.00s)2.

The solution to this quadratic equation

And find t

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