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sha0403
a ball thrown straight up into the air so that its height in feet after t seconds is given by s(t)= 128t-16t^2.. so, find the average velocity of the ball during tie interval (2,2.1)
yes.. after that i get 128-32t.. so what should i do that given time interval (2, 2.1)
U need to find {s'(2) + s'(2.1)}/2
why it divide by 2? can u explain to me?
Because we are calculating avereage of two velocities
oh like that..if we want to calculate average between 3 velocity,, we should divide by 3 also?
add three valocities and dvide it by
ok.... i would like to ask u something.. if the question ask to find instantaneous velocity at t=2.. what the mean of instantaneous/
instantaneous means velocity of that body at that particular time
Umm, one thing I'd like to note is that @sauravshakya 's method for finding the average value works for lines. In this case velocity is indeed linear so it will work, though.
Wait, actually @sauravshakya , the units don't seem to work out with your method. You have \(\frac{m/s}{s}\), which is acceleration.
I think we don't need to do any calculus at all here. It's just: \[ \large \frac{s(t_f)-s(t_i)}{t_f-t_i} \]
I was just calculating average velocity
Wow, that's right. I was wrong.
@wio your formula when i should use it? what the kind of question?
Anyway, I'm thinking that the average velocity, given velocity is: \[ \large \overline{v(t)} = \frac{1}{t_f-t_i}\int_{t_i}^{t_f} v(t) dt \]Note that \(s(t) = \int v(t) dt \), which gives us (by fundamental theorem of calc: \[ \frac{s(t_f) - s(t_i)}{t_f - t_i} \]
@sauravshakya and @wio ..thank u very much