Find displacement and distance traveled

Question

The velocity function is v(t) = t^2-5t+6 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,6]

displacement=

distance traveled =

Answer 1

To find the displacement you would integrate v(t) from 0 to 6.

∫v(t)dt = ∫(t^2-5t+6)dt = t^3/3 - 5/2 t^2 +6t +C

So the definite integral from 0 to 6 (aka the total displacement) is (6^3)/3 -5/2 (6^2) +6*6 - (0/3 + 0/2 +6*0) = 18 units.

For the distance traveled, you need to know when the velocity is positive and negative.

Since the graph of v is a parabola with positive leading coefficient and positive constant term, you know the velocity starts positive. Now you need to find out where the velocity is 0.

Setting v(t) = t^2-5t+6 = (t-2)(t-3) = 0, you know this happens when t = 2 or 3. From this and the shape of the graph, you know it's positive on [0,2) and (3,6] and negative on (2,3). So the total distance traveled is ∫v(t)dt from 0 to 2, plus ∫v(t)dt from 3 to 6, plus the absolute value of the integral ∫v(t)dt from 2 to 3.

∫v(t)dt from 0 to 2 = (2^3)/3 -5/2 (2^2) +6*2 - (0/3 + 0/2 +6*0) = 14/3 - 0 = 14/3

∫v(t)dt from 2 to 3 = (3^3)/3 -5/2 (3^2) +6*3 - ((2^3)/3 -5/2 (2^2) +6*2) = 9/2 - 14/3 = -1/6

∫v(t)dt from 3 to 6 = (6^3)/3 -5/2 (6^2) +6*6 - ((3^3)/3 -5/2 (3^2) +6*3) = 18 - 9/2 = 27/2

So the total distance traveled is 14/3 + 1/6 + 27/2 = 55/3 units.

Answer 2

v(t)=t^2-5t +6, from 0 to 6

when v = 0, the particle changes direction

v=0 = (t-3)(t-2)

at time t=2 and t=3

it goes forward from 0 to 2, backwards from 2 to 3, then forwards from 3 to 6

s(t) = displacement at time t

s(t) = the integral of v(t)

= t^3/3 -5t^2/2 +6t + c

let c=0

at time t=0, the particle is at point 0

t=2, at 4 2/3

t=3 at 4 1/2

t=6 at 18

it moved from 0 to 18

displacement =18

but

movement was 18+2(4 2/3 - 4 1/2) = 18+ 2(1/6)

= 18 1/3 = total distance traveled

it traveled the same distance 3 times from t=2 to 3

back and forth