Using standard eq. v=s/t can a vehicle accelerate to a speed of 114m/hr in a distance of .9tenth of a mile.

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- Feb 25th 2008, 01:29 PMcst301[SOLVED] speed & velocity
Using standard eq. v=s/t can a vehicle accelerate to a speed of 114m/hr in a distance of .9tenth of a mile.

- Feb 27th 2008, 12:13 PMAryth
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.