Calculus
posted by D on .
A particle moves in a straight line with velocity t^2  1/9 ft/s. Find the total displacement and total distance traveled over the time interval [1,4].
I found out that the total displacement is .4116
But I cannot find the total distance traveled.

Let
v(t)=velocity function = t^2  1/9
s(t)=displacement function. = t/91/t
and
displacement = s(4)s(1)= 25/36  (10/9) = 5/12 ft.
If s(t) is monotonically increasing or decreasing, then the displacement equals the distance travelled.
However, we note that v(3)=0, after which time the velocity reverses in direction.
So the distance travelled equals
s(4)s(3)  [s(3)s(1)]
=25/36 (2/3)  [(2/3)(10/9)]
=17/36
Ignore the sign, since distance is a scalar.
So distance = 17/36 ft.