Asked by D
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.
I found out that the total displacement is .4116
But I cannot find the total distance traveled.
Answers
Answered by
MathMate
Let
v(t)=velocity function = t^-2 - 1/9
s(t)=displacement function. = -t/9-1/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.
v(t)=velocity function = t^-2 - 1/9
s(t)=displacement function. = -t/9-1/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.
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