Can Police Catch the Speeding Motorist?

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The discussion revolves around calculating the distance at which police can catch up to a speeding motorist. After a 2-second delay, the police accelerate for 15 seconds, covering 281.25 meters and reaching a speed of 37.5 m/s. Meanwhile, the motorist travels 510 meters in the same time frame, leaving the police 228.75 meters behind. With a relative speed of 7.5 m/s, it takes an additional 30.5 seconds for the police to catch up, totaling 47.5 seconds from the initial observation. The intersection occurs 1,187.5 meters from the observation point.
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How would I solve this?

2 police officers are standing on a road. A motorist comes at 108km/hr(30m/s) The police take 2 s to assess the situation. Then their patrol car accelerates at 2.5m/s^2 for 15s after the 2s and then continues at uniform speed. How far from the observation point does the intersection take place? Assume the motorist maintains a constant speed of 108km/hr?
 
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Yes you do, assuming the question means the police react as the bike passes parallel to them.

using:

S = ut + 1/2*a*t2

You find that the police travel 281.25m in the 15 seconds they accelerate, plus the 2 seconds they waited, so in 17s they have traveled 281.25m and accelerated to a constant velocity of 37.5m/s

In this time the motorbike has traveled 17*30 = 510m so the police after 17s are a distance of 228.75m behind the bike

The relative speed between the two is 7.5m/s

So how long does it take to make up 228.75m at 7.5 m/s?

228.75 / 7.5 = 30.5s

plus the original 17s

30.5 + 17 = 47.5s until the police are level with the biker.
 
Thank you ,but I still am not sure of the answer to "how far from the observation point did the intersection take place?"
 
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