Question

Intro:

On July 25, 2010, a police officer was sitting in his police cruiser adjacent to the 21st street bridge on U.S. 6, in Big County, Ohio. The police officer was facing east watching traffic. At around 3 AM, the police officer saw headlights approach at high speed. He activated his radar and clocked the approaching suspect car at 112 miles per hour. After the suspect car had passed the officer's location, the officer turned left and proceeded to follow the suspect. The officer turned his lights and siren on and attempted to catch up with the suspect.

Goal: We want to calculate the distance between the officer and the suspect when the suspect exited the highway at the Miami Grove exit.

To do so we make further assumptions: Data from Michigan State Police Vehicle Evaluation is used to model the police car's speed (data is from tests performed on a 2009 4.6L Ford Police Interceptor). It is .928 mi (4900 feet), travelling on US 6, from the 21st street bridge to the beginning of the Miami Grove exit ramp. The suspect maintained his speed of 112 miles per hour slowing down to 80 miles per hour just as he reached the Miami Grove exit ramp.

1. Velocity function of the suspect and his total time travelled on the highway.

As the officer clocked the suspect at 112 mi/hr or 164.2667 ft/ sec. we assume the client continued at this speed until slowing to exit just before the exit ramp. We assume his exit speed is 80mi/hr or 117.3333 ft/sec. Also, assume that he decelerated at a constant rate for the last 300 feet of his trip on the highway. Since the total distance is 4900 ft, we see that it took him 28 seconds to drive 4600 ft:

164.2667 ft/sec x 28 sec = 4599.47 ft ~ 4600 ft

The suspect's velocity function than for the first 28 seconds of the chase is simply the constant value of 167.2667 ft/sec. We want to find his velocity as a function of time for the last 300 ft of his highway trip.

Find A, B, and Texit in the following piece-wise function for the suspect's velocity: v(t) _c(t)

167.2667 for0 < 28 t < (t) IA(t– 28) + B 28 <t < Terit .

where A and B are constants and T_exit is the time the suspect exited the highway

2. What was the distance between the officer and the suspect when the suspect exited the highway?

Answer 1

Here First 4600 ft out of 4900 ft is covered at a speed of 167.2667 ft/sec for 28 sec.

The last 300 ft is covered at a speed of 117.3333 ft/sec ,

so, the exit time for the above 300 ft distance = 300/117.3333 sec = 2.557 sec

So, Texit = 2.557 sec

Since from the velocity function we get,

$v(t)=A(t -28 )+B$    FOR $28\leq t\leq Texit$

so, at t =28 sec, v(t) = 167.2667 ft/sec (given)

so, we get B = 167.2667 FT/SEC

And, at Texit= 2.557 sec,v(t)=117.3333ft/sec

we get,

A=1.96255 ft/sec²

From $v(t)=A(t -28 )+B$ after integration we get,

  $r(t)=A(\frac{t^2}{2}-28t)+Bt$

Where r(t) tells the position between the officer and the suspect.

so, the distance bettween officer and suspect at the excited state i.e, When Texit= 2.557 sec,

= 1.96255((2.557²/2)-(28*2.557))+(167.2667*2.557) = 293.606 ft