For an object whose velocity in ft/sec is given by v(t) = -3t2 + 6, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

Question

A. 4

B. 7.314

C. 3.657

D. -6

Answer 1

Not clear what you mean my "distance"

integrating:

x(t) = -t^3 + 6t <---- assumes the initial position at t=0 is zero, so no constant

x(2) = -8 + 12 = 4 ---> (2,4)

x(0) = 0 ---> (0,0)

the naive answer is x(2)-x(0) = 4-0 = 4 is the 1-d distance

distance formula says sqrt( 2^2 + 4^2) = sqrt( 4 + 16) = sqrt(20) = 2 * sqrt(5) which is not one of the

choices

If you want to find the arc length, then you have to integrate

integral sqrt( 1 + (6-3t^2)^2) =

integral sqrt( 1 + 9 ( 2 - t^2)^2)

good luck with that, probably have to do it numerically using simpson's rule

WolframAlpha says 7.724266 which is also not an answer

pleas repost

Answer 2

Distanse S = ∫₀²abs(-3t² + 6)dt;

-3t² + 6 ≥ 0 for t² ≤ 2, 0 ≤ t ≤ 2^1/2;

S = ∫₀^{2^(1/2)}(-3t² + 6)dt + ∫_2^(1/2)²(3t² -6)dt= (- t³ + 6t)₀^{2^(1/2)} + (6t - t³)_2^(1/2)² = - 2^3/2 + 6 · 2^1/2 + 12 - 8 - 6 · 2^1/2 + 2^3/2 = 4;

Answer: distance S = 4