# constant acceleration problem

• Aug 30th 2009, 01:54 PM
suboba
constant acceleration problem
hi,

I'm having a bit of trouble with the concept of constant acceleration.

If a car goes from 0 to 80 mph in six seconds with constant acceleration, what is that acceleration?

My initial thought was to take the change in velocity (80mph) and divide it by the change in time (6s). But I'm not really sure what constant acceleration means.

the answer should be in ft/sec^2. I tried converting everything to ft and sec, but I still couldn't get the right answer. Plz help . thank you.
• Aug 30th 2009, 02:01 PM
e^(i*pi)
Quote:

Originally Posted by suboba
hi,

I'm having a bit of trouble with the concept of constant acceleration.

If a car goes from 0 to 80 mph in six seconds with constant acceleration, what is that acceleration?

My initial thought was to take the change in velocity (80mph) and divide it by the change in time (6s). But I'm not really sure what constant acceleration means.

the answer should be in ft/sec^2. I tried converting everything to ft and sec, but I still couldn't get the right answer. Plz help . thank you.

Convert 80mph into ft/s

1 mile = 5280ft

1 hour = 3600s

1mph = 5280/3600 ft/s

Once it's in ft/s divide by 6 to get the answer
• Aug 30th 2009, 02:13 PM
skeeter
good conversion factor to remember ...

$60 mph = 88 ft/sec$
• Aug 30th 2009, 02:14 PM
eXist
I usually think of acceleration as such:

Let's say you're given a standard equaiton: $f(x)$
I'm going to let $f(x)$ represent a position equation I can represent as $s(x)$.

So at the moment we have $f(x) = s(x)$
Lets take the derivative of the two:
$\frac{d}{dx}f(x) = \frac{d}{dx}s(x)$
$f'(x) = s'(x)$

However, most people will call $s'(x)$ the velocity function of $f(x)$ so I'll let $v(x) = s'(x)$:
$f'(x) = v(x)$

Finally we repeat this step again and we have:
$\frac{d}{dx}f'(x) = \frac{d}{dx}v(x)$
$f''(x) = v'(x)$

Once again I'm going to let $a(x) = v'(x)$, so:
$f''(x) = a(x)$

This $a(x)$ we denote as acceleration.

So here is the whole thing together:
$f(x) = s(x) \Rightarrow$ s(x) being the position function (original function)
$f'(x) = v(x) \Rightarrow$ v(x) being the velocity function (first derivative)
$f''(x) = a(x) \Rightarrow$ a(x) being the acceleration function (second derivative)