The velocity function is v(t)= -(t^2)+6t-8for a particle moving along a line.Find the displacement and the distance traveled by the particle during the time interval [-2,5].

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that you may find the displacement evaluating the definite integral of velocity function, such that:

`d(t) = int_(-2)^5 (-(t^2) + 6t - 8) dt`

You need to use the property of linearity of integral such that:

`d(t) = int_(-2)^5 (-(t^2)) dt + int_(-2)^5 6t dt - int_(-2)^5 8dt`

You need to use the fundamental theorem of calculus, such that:

`d(t) = -t^3/3|_(-2)^5 + 6t^2/2|_(-2)^5 - 8t|_(-2)^5`

`d(t) = -5^3/3 + (-2)^3/3 + 3*25 - 3*4 - 8*5 - 8*2`

`d(t) = -133/3 + 11 => d(t) = (-133+33)/3 = -100/3`

You need to evaluate the distance, hence, you need to find where the function of velocity is positive and negative, hence, you need to find its zeroes such that:

`-t^2 + 6t - 8 = 0 => t^2- 6t+ 8 = 0`

You need to use the quadratic formula such that:

`t_(1,2) = (6+-sqrt(36 - 32))/2 => t_(1,2) = (6+-2)/2`

`t_1 = 4 ; t_2 = 2`

Notice that the function is positive over the intervals [-2,2] and [4,5] and it is negative over the interval [2,4] such that:

`|t^2 - 6t + 8| = (t^2 - 6t + 8, t in [-2,2] U [4,5]) or (-t^2 + 6t - 8, t in [2,4])`

`s(t) = int_(-2)^2 (t^2 - 6t + 8) dt + int_2^4 (-t^2 + 6t - 8) dt + int_4^5 (t^2 - 6t + 8) dt`

`s(t) = (t^3/3 - 3t^2 + 8t)|_(-2)^2 + (-t^2 + 6t - 8t)|_2^4 + (t^2 - 6t + 8t)|_4^5`

`s(t) = 8/3 - 12+ 16 + 8/3 + 12 + 16 - 16 + 24 - 32 + 4 - 12 + 16 + 25 - 30 + 40 - 16 + 24 - 32`

`s(t) = 16/3 + 27 => s(t) = 97/3`

Hence, evaluating the displacement yields `d(t) =-100/3`  and the distance yields  `s(t) = 97/3` .