yea i got 14.8 seconds as well, but i originally simplified the problem using relativity. the speeder can be assumed to be going 50km/h while the cop is stationary. 50km/h = 13.89 m/s.
I divided the problem into 3 parts:
1) after 1 second
2) when cop speed = 50km/h (cop and speeder are going at same speed)
3) when cop finally catches up to him
so when we do the math at the end, we get t1 + t2 +t3 = t(total) ], which is our answer.
(FOR THIS PROBLEM I WILL BE USING M/S UNTIL THE VERY END, SO I WONT WRITE IT ALL THE TIME) (v is implied in m/s unless otherwise stated)
1.)after one second
cop: no displacement because stationary, vi = vf = 0, t = 1
speeder: displacement = vt; 13.89m = (13.89m/s)(1s)
***
1 second has passed
2.) when cop speed = 50km/h (cop and speeder are going at same speed)
cop: vi = 0, vf = 13.89, a = 2
using UAM equation: vf = vi + at, 13.89 = 0 + 2t, t=6.945
displacement = [(vi+vf)/2](t), displacement = [(0+13.89)/2](6.945) = 48.23
speeder: vi = 13.89, vf = 13.89 t = 6.945
using v = displacement/time, 13.89 = displacement/6.945, displacement = 96.47
cop has travelled 48.23m , speeder has travelled 96.47m, their speeds are equal
***
6.945 seconds have passed JUST in this part
the cop is (96.47-48.23) = 48.24m away from the speeder
they both have a speed of 13.89
3) when cop finally catches up to him
cop: displacement = vi(t) + 1/2(a)(t2), displacement + 48.24 = 13.89t + 1/2(2)(t2)
*(the +48.24 next to displacement in this equation is to account for the distance between the cop and the speeder at this point in time. to catch the speeder, the cop will have to travel the same distance as the speeder in this section PLUS make up for the gap that was already existing (48.24m))
speeder: displacement = vi(t) +1/2(a)(t2), displacement = 13.89(t) + 1/2(0)(t2), displacement = 13.89t
(i am aware that displacement = v(t) for a constant velocity object, but i just used a UAM equation and took the long route to create continuity)
setting displacements equal to each other:
13.89t + 1/2(2)(t2) - 48.24 = 13.89t
*keep in mind that "t" in this part is NOT the final answer. it is only referring to the time that it takes the cop to finally catch the criminal after the previous 2 parts have occured (1 second passed and cop accelerates until his speed = speeder speed). It may be easier to think of t in this part as t3. I didn't use t1 t2 or t3 because it can be easily mistaken as a coefficient when not in subscript)
subtracting 13.89t from both sides, and moving 48.24, and simplifying:
t2 = 48.24
t= 6.945
interesting! t3 = t2
So now we finally solve:
t1 + t2 + t3 = t(total)
1 + 6.945 + 6.945 = t(total)
t(total) = 14.89 seconds
it takes 14.89 seconds for the cop to catch the speeder
