Solve for time, given distance and acceleration

Question

Two gangsters are flying down I-70 West at a constant 108 hm/hr. If they make it to Indiana, they will be safe. When they are 1 km away from the Indiana border, while still traveling at a constant 108 km/hr, they pass a concealed police car hidden in a speed trap. At the instant that the criminals pass the patrol car, the cop pulls onto the highway and accelerates at a constant rate of 2 m/s^2. Does the cop catch up with them before they cross the state line?

Here's what I have progressed so far

ΔV = (ΔX)/(T)

when V is velocity, X is distance, and T is time.

108 = (1 - 0)/T

108T = 1

T = 1/108

T = 0.00925 hours

T = 0.00925 * 60 * 60

T = 33.33 seconds (this is the time it takes the gangsters to get to the borders of Indiana)

What I am stuck at is the time it takes the cop to travel the same distance. I have used the distance formula to find time as follows:

Xf = Xi + 1/2 * A * T^2

where A is acceleration

Since the cop is starting from rest, the initial velocity is 0 (zero), so

1 = 0 + 1/2 * (2) * T^2

1 = T^2

T = 1 seconds

So it takes the cop 1 seconds to travel the whole mile to Indiana? What am I doing wrong?

Answer 1

The acceleration is in meters/second^2 but the distance is 1 kilometer=1000 meters. The left side should be 1000, not 1.

Answer 2

Converting the gangsters' speed to ms$^{-1}$, $108$ km/h $=108/3.6$ ms$^{-1} = 30$ ms$^{-1}$.

The distance the cop travelled in time $t$ s is $\frac12at^2 = t^2$ metres, and the distance the gangsters travelled in time $t$ s is $30t$ metres. The time when they met the second time is $$\begin{align*}t^2 =& 30t\\t=&30\text{ seconds}\end{align*}$$

Then the distance they have travelled in $30$ seconds is $900$ metres, and this is before they arrive Indiana.