Average acceleration over interval

let's say that we have a particle that's traveling in one dimension and its position as a function of time is given as T to the third power plus two over T squared and what I would like you to do is pause this video and figure out what the average acceleration is of this particle over the interval the closed interval from T is equal to 1 to T is equal to 2 what is this what is this going to be equal to so assuming you've given a go at it and the first thing you might have realized is we're trying to take the average value of a function that we don't know explicitly yet we know the position function but not the acceleration function but luckily we also know that the acceleration function is the derivative with respect to time of the velocity function which is the derivative with respect to time of the position function so the acceleration function is the second derivative of this and then we have to just find its average value over this interval so let's do that let's take the derivative of this twice and before we do it let me just even rewrite this so it's going to be a little bit easier to differentiate it so if we just take each of these two terms of the numerator and divide them by T squared we're going to get T to the third divided by T squared is just T and then 2 divided by T squared we could write that as plus 2t to the negative 2 power and now let's let's take the derivative so the velocity function as the velocity as a function of time just a derivative of this with respect to time so it's going to be derivative of T with respect to T is 1 derivative of 2t to the negative 2 let's see negative 2 times positive 2 is negative 4t to the and we just decrement the exponent here T to the negative negative 3 power now to find acceleration as a function of time we just find take the derivative of this with respect to time so acceleration as a function of time is equal to actually sits already use that color for the average let me do a different color now so acceleration as a function of time is just the derivative of this with respect to T so derivative of a constant with respect to time what's not changing so it's zero and then over here negative 3 times negative or is positive 12 times T to the let's decrement that exponent to the negative 4 power now to find the average value all we have to do now average average value is essentially take the definite integral of this over the interval and divide that by the width of the interval so or we could say we could take we can divide by the width of the interval 1 over 2 minus 1 and this all simplifies to 1 times the definite integral over the interval so 1 to 2 of a ft which is so this can be 12 T to the negative 4 power D T so what does this simplify to once again this is 1 over 1 that's just going to be 1 if we take the anti derivative of this and then we actually well let me just so this is going to be equal to the antiderivative this is so we're going to go t to the negative 3 power but then we divide by negative 3 so an antiderivative of this is going to be if we don't take the well an antiderivative is going to be negative 4 T to the negative 3 power and we saw that over here obviously if you were really just taking an indefinite integral we would have to put some constant here but in the definite integral even if we put a constant here it would get can't if we if we assuming the same constant it would get cancelled out when you actually do the calculation but the antiderivative of this we increment the exponent and then we divide by that new exponent so 12 divided by negative 3 is negative 4 and we're going to evaluate that if I am at 2 and at 1 and so this is going to be equal to when we evaluate it at 2 at the upper bound of our interval it's going to be negative 4 times 2 to the negative 3 power so it's negative 4 times what is that - that's 1 over 2 to the 3rd over times 1/8 is one way to think about that and then we're going to have minus this evaluate at 1 so minus negative 4 times T to the negative 3 1 to the negative 3 is just 1 so it's just going to be negative 4 times 1 and this is going to be equal to or really in the homestretch now this is equal to this part right over here is negative 1/2 so this is negative 1/2 and this part right over here is positive 4 so positive 4 minus 1/2 we could either write that as 3 and 1/2 or if we wanted to write it as an improper fraction we could write this as 7 halves so the average value of our acceleration over this interval is 7 halves and if the position was given in meters and time was in seconds then this would be 7 halves meters per second squared is the average acceleration between time and one second and time at two seconds