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),
so1 = 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?
Best Answer
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.