bullet

Police Car Catching with Speeding Motorist

A police car is at the side of a 65 Mi/H speed limit highway when a motorist drive by at 76 Mi/H. The police pursues the motorist until the two vehicles are driven parallel to each other in two adjacent lanes. If the police car accelerates uniformly to the top speed of 85 Mi/H at a rate that correspond to an increment in velocity of 6 Mi/H per each second of acceleration,

How long does it take the police to catch with the motorist?

N

a)

 67 s

 

b)

 72 s

 

c)

 144 s

 

d)

 168 s

 

e)

None of the above.

How far, from the point where the motorist drove by the police car, does the police catch with the driver?

 

a)

 2.8 Km

 

b)

 1.7 Km

N

c)

 2.3 Km

 

d)

 3.6 Km

 

e)

None of the above.

Solution:

The motion can be divided in two parts, the motion when the police car is accelerating and the motion when the police car moves at constant velocity. At the same time, the motorist moves at constant velocity along the entire motion. The first drawing of the applet shows the beginning of the motion, when the police car is at rest and accelerating at the rate of ap = 6 Mi/H per second. The motorist moves at vc = 76 Mi/H all the time. The second drawing of the applet shows the instant when the police car reaches the top velocity of vp = 85 Mi/H after traveling the distance x1. The last drawing of the applet shows the instant when the police car reaches the motorist. This is achieved after the police car travels the additional displacement  x2. Thus, the police car catch the motorist after traveling the distance x = x1 + x2. This is the same distance that the motorist travels at constant velocity. The time needed by the police to catch with the motorist can also be divided accordingly with the two parts of the motion, t = t1 + t2.

Police Motion:

First part, motion with constant acceleration starting from rest,

Time taken, t1. Equation (U_A_M 3) with original velocity zero.
Displacement x1. Equation (U_A_M 4) with original velocity zero. Substituting the previous calculated time,

Second part, motion with constant velocity,

Displacement x2. This equation is obtained from the definition of velocity for motion with constant velocity.

From the previous two parts of the motion, the total displacement is

Motorist Motion:

Displacement x. This equation is also obtained from the definition of velocity for motion with constant velocity.
Time taken, t. Where the previous result for the time t1 has been substituted.

Therefore, the total displacement is also

Combination of the two Motions:

The two total displacement of the police and the displacement of the motorist are the same. Thus, equating the two previous expressions for the displacement an equation for the time t2 is obtained,

The applet on the left shows details of the algebraic manipulations of the formulas.

Some general results can be concluded from the last expression for the time  t2 corresponding to the police car moving with constant velocity. First when the final velocity of the police car is the same as the velocity of the motorist, this time becomes very large; in fact, the police will never catch the motorist. On the other side, if the police car reaches a maximum velocity equals to twice the velocity of the motorist, the police car catches the motorist just when it stops accelerating. These conclusions are not necessary for solving the problem.

Still the total time is not jet calculated,

After the time has been calculated, the total displacement can be easily obtained by just substituting this time into the equation of motion for the motorist because this motion is with constant velocity. Therefore,

Numerical calculation:

Thus, the total time is t = 67 s and the total displacement is x = 2.3 Km.

 

bullet
by Luis F. Sez, Ph. D.    Comments and Suggestions: LSaez@dallaswinwin.com