[SOLVED] speed & velocity

cst301 · February 25th 2008, 01:29 PM

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

**Aryth** · February 27th 2008, 12:13 PM

Well of course it's possible.

First we have the following information:

$\text{Initial Speed }= 0 \frac{m}{hr}$
$\text{Final Speed }= 114 \frac{m}{hr}$
$\text{Distance }= 0.9 m$

According to the standard equation:

$v = \frac{s}{t}$

Plug-n-chug:

$114 = \frac{0.9}{t}$

$114t = 0.9$

$t = \frac{0.9}{114}$

$t = 0.008 hrs$

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 $\frac{m}{hr}$ so we have another standard equation:

$a = \frac{v}{t}$

It's similar to the speed formula, but it is changed around for acceleration:

$a = \frac{114}{0.008}$

$a = 14250 \frac{m}{hr^2}$

That is the acceleration you must have in order to attain the velocity in the time allotted over the distance allotted.

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