Using standard eq. v=s/t can a vehicle accelerate to a speed of 114m/hr in a distance of .9tenth of a mile.
Well of course it's possible.
First we have the following information:
$\displaystyle \text{Initial Speed }= 0 \frac{m}{hr}$
$\displaystyle \text{Final Speed }= 114 \frac{m}{hr}$
$\displaystyle \text{Distance }= 0.9 m$
According to the standard equation:
$\displaystyle v = \frac{s}{t}$
Plug-n-chug:
$\displaystyle 114 = \frac{0.9}{t}$
$\displaystyle 114t = 0.9$
$\displaystyle t = \frac{0.9}{114}$
$\displaystyle t = 0.008 hrs$
It is possible to accelerate to that speed, in fact, it would take the same amount of time as specified above.
Now, we know the change in velocity was 114 $\displaystyle \frac{m}{hr}$ so we have another standard equation:
$\displaystyle a = \frac{v}{t}$
It's similar to the speed formula, but it is changed around for acceleration:
$\displaystyle a = \frac{114}{0.008}$
$\displaystyle a = 14250 \frac{m}{hr^2}$
That is the acceleration you must have in order to attain the velocity in the time allotted over the distance allotted.