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 a_{p}
= 6 Mi/H per second. The motorist moves at
v_{c}
= 76 Mi/H all the time. The second drawing of the applet shows
the instant when the police car reaches the top velocity of
v_{p}
= 85 Mi/H after traveling the distance
x_{1}. 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 x_{2}.
Thus, the police car catch the motorist after traveling the distance
x =
x_{1} + x_{2}.
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 =
t_{1} + t_{2}.
Police Motion:
First part, motion with constant acceleration
starting from rest,
Time taken, t_{1}. 

Equation (U_A_M 3)
with original velocity zero. 

Displacement x_{1}. 

Equation (U_A_M 4)
with original velocity zero. Substituting the previous
calculated time, 

Second part, motion with constant velocity,
Displacement x_{2}. 

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
t_{1} 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
t_{2} 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
t_{2}
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. 