For an object whose velocity in ft/sec is given by v(t) = −2t2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

Question

For an object whose velocity in ft/sec is given by v(t) = −2t² + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

 A.4.876

B.  2.667 

C. 9.752

 D. 2.438

Answer 1

Distance s = ∫₀²I-2t² + 4Idt = ∫₀^√2(4 - 2t²)dt + ∫_√2²(2t² - 4)dt = (4t - 2t³/3)₀^√2 + (2t³/3 - 4t)_√2² =

= 4√2 - 4√2/3 + 16/3 - 8 - 4√2/3 + 4√2 = 8√2 - 8√2/3 - 8/3 = (16√2 - 8)/3 = 4.876

Answer is A.

Answer 2

s(t) = the integral of v(t) = -2t^3/3 + 4t

evaluate at t=2

-2(8)/3 + 4(2) = -16/3 + 8 = 8-5 1/3 = 2 2/3

B is correct rounded off to 3 decimal places, exact answer is 2.666.... where the 6's never end