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Acceleration = rate of change of velocity = distance units / time units squared
So, it could be ft/sec^2, inches/hour^2, or any such thing.
it doesn't have to be specifically m/s^2 like GT , it is going to be distance over time^2 so it can be ft/s^2 or anything
oh i see... i have web homework..i was putting m/s^2 and it wasnt right..maybe i should try ft/sec^2
well it depends on the question, if you show us we can help. it won't necessarily be ft/s^2 but it will depend on the way your question is worded
Suppose that an accelerating car goes from 0 mph to 68.2 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 ft/sec.) Find the average acceleration of the car over each of the first two seconds. t (s) 0 1 2 3 4 5 v(t) (ft/s) 0.00 34.09 59.09 77.27 90.91 100.00 give the average acceleration over the 1st and 2nd second 1st second ___ = 2nd second ___ =
It is the difference in velocities during those periods divided by the time elapsed.
it will be units of ft/s^2 like you had thought!
man calc 1 here at this school is TOUGH!!!
NOT this problem persay...but the exams
and in my head, i think your answer should be 25 ft/s^2
that question wasn't really calculus aha.
and calculus is never easy until you fully understand it! it required much deeper thought than regular math
haha i mean, overall calc here is hard...and umm yea i think the answer was 25 ft/sec^2 or something lol