Thread: Integral

The velocity function (in meters per second) is given for a particle moving along a line.
a. Find displacement
b. Distance traveled.

V(t)= 3t -5 , 0 <= t <= 3

Originally Posted by camherokid

The velocity function (in meters per second) is given for a particle moving along a line.
a. Find displacement
b. Distance traveled.

V(t)= 3t -5 , 0 <= t <= 3

Here

Originally Posted by Jhevon

Here

answer for
a. -3/2 m
b. 41/6 m

Originally Posted by camherokid

answer for
a. -3/2 m
b. 41/6 m

ok, so apparently i mixed up the questions, i got the answer for (a)

the answer was negative since the part of the curve was under the x-axis (or t-axis) was larger than the part above it. for distance travelled however, we cannot have a negative value. so we must find these two areas in terms of magnitude and add them together. to find the areas as positive maginitudes i needed to split the integral in 2 and turn the limits of the first backward. that gives me positive areas for both pieces

Here's the graph so you see what i'm talking about:

Hello, camherokid!

the velocity function (in m/sec) is given for a particle moving along a line.
. . v(t) .= .3t - 5, .0 < t < 3

(a) Find displacement
(b) Distance traveled

The position function is: .s(t) .= .∫(3t - 5) dt .= .3t²/2 - 5t + C


(a) When t = 0: .s(0) .= .3·0²/2 - 5·0 + C .= .C

. . .When t = 3: .s(3) .= .3·3²/2 - 5·3 + C .= .C - 3/2

The displacement is: .(C - 3/2) - C .= .-3/2

(The particle is 1½ units to the left of where it started.)


(b) The velocity is zero when t = 5/3
. . .(That's when the particle changes direction.)

We know that: .s(0) = C

When t = 5/3: .s(5/3) .= .C - 25/6
. . The displacement is: .(C - 25/6) - C .= .-25/6
It has moved 25/6 units to the left.

When t = 3: .s(3) .= . C - 3/2
. . The displacement is: .(C - 3/2) - (C - 25/6) .= .8/3
It has moved 8/3 units to the right.

Therefore, the total distance is: .25/6 + 8/3 .= .41/6 units.