How do I find the distance traveled if I know the acceleration and time?

[vbman11]

so if A is the acceleration
and T is time
and D is distance
then

D = A * T^2

right?

 

[berkeman]

so if A is the acceleration
and T is time
and D is distance
then

D = A * T^2

right?

Here's a good intro for you:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

.

 

[vbman11]

Here's a good intro for you:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

.

I was looking for a quick check to make sure I'm right... so yes or no?

 

[berkeman]

I was looking for a quick check to make sure I'm right... so yes or no?

Please look at the link. The answer is right there in a big box on the front page.

 

[vbman11]

Sry bud you just sent me a link to a page that contains an HTML frame. go ahead and try that link out for yourself and see. It'll just show the home page.

 

[berkeman]

Sry bud you just sent me a link to a page that contains an HTML frame. go ahead and try that link out for yourself and see. It'll just show the home page.

Oopsies. I didn't realize the link didn't follow me down a couple of click levels. I'll have to figure out how to fix that.

In the mean time, just click on Mechanics, and then on Velocity & Acceleration.

Does that work okay?

 

[litup]

so if A is the acceleration
and T is time
and D is distance
then

D = A * T^2

right?

Nope, it's (A *T^2 )/2

So 9 seconds and 32 f/s (1g) would cover about one quarter mile, 1320 feet.
32*81=2592, divide by 2=1296 feet. A true quarter mile would be about 9.08 seconds.

73 Berkeman, AI3N here. 2 meters, 70 cm, 20/30/40/80 voice mainly. Antenna blew down, need one of those air powered potato guns to get the dipole back in the trees!