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An object falls because of gravity at a rate of -980 cm/sec2 (- sign is indicative of a downward direction) for 10 seconds. If it starts from rest (vo= 0) and its position starts from do= 0, where is it 10 seconds later?
_______________m
I don't know where to start
maybe use d=1/2vt but what do I plug in
arildno
Nov22-04, 11:23 AM
maybe use d=1/2vt but what do I plug in
Whatever is this??
You have constant acceleration, how is distance as a function of time given in that case?
tony873004
Nov22-04, 01:43 PM
If every second, it is moving 980 cm / second faster, then how fast is it going in 10 seconds?
v = a * t
what is its average velocity during this 10 second peroid? Hint: It started at 0, it ended at a*t. Take the average.
Now distance is easy. It travelled an average velocity for a known time.
d=1/2vt
This formula is wrong.
d = vt
Following the logic above, you should be able to come up with the correct formula for d using a (acceleration) instead of v (veolcity).
I think it is 980m
This is wrong too. How did you get this? Watch your units. You expressed acceleration in cm, but your answer in meters, which is ok to do, but make sure you convert properly.
Just use this formula
\Delta x = \frac{1}{2}at^2
Input the values and solve. -4.9m(100s)
-490 meters from its original position
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