Basic Kinematics problem, why my method is invalid

[Yousufshad]

Homework Statement

A motorist drives along a straight road at a constant speed of 14.4m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 1.8m/s^2 to overtake her. Assuming the officer maintains this acceleration, determine the time it takes the police officer to reach the motorist (in seconds).

Homework Equations

vi^2 + 2ad = vf^2
v=at
d=vt

The Attempt at a Solution

Here is what I did I am curious why my method is incorrect.
(14.4)t=d (speeder)
vt=d (officer)
v=at (officer)
at^2=d (officer)

1.8t^2 = 14.4 t
t=0, 8

Correct answer is 16

Ok I realized that d = 1/2 vt^2
this was my mistake, please show mathematically why I can't use these two equations to form d =vt^2

vt=d
vf =at (if vi=0)
(at)t=d
at^2 =d

 

[gneill]

Ok I realized that d = 1/2 vt^2
this was my mistake, please show mathematically why I can't use these two equations to form d =vt^2

vt=d
vf =at (if vi=0)
(at)t=d
at^2 =d

vt = d only if velocity is constant over time. In this case it is not since there is acceleration involved. So you must turn to the full kinematic expression:

$d(t) = d_o + v_o t + \frac{1}{2} a t^2$

 

[Yousufshad]

vt = d only if velocity is constant over time. In this case it is not since there is acceleration involved. So you must turn to the full kinematic expression:

$d(t) = d_o + v_o t + \frac{1}{2} a t^2$

Ok, thanks, great to know :)