Rina asked in Science & MathematicsPhysics · 1 month ago

The velocity is v(t)=−t2+6t−8 for a particle moving along a line. Find the the distance traveled during the time interval [-3,6].?

4 Answers

Relevance
  • NCS
    Lv 7
    1 month ago

    I'll show you how to do what @andrew suggested.

    Find where the velocity changes sign:

    v = 0 = -t² + 6t - 8

    This is quadratic with roots at t = 2 and t = 4

    At t = 3, v = -3² + 6*3 - 8 = 1

    which is positive

    and so the velocity must be negative for t < 2 and t > 4.

    s(t) = ∫ v(t) dt = -t³/3 + 3t² - 8t + C

    Evaluated for -3 < t < 2, s = -66.7

    and for 2 < t < 4, s = 1.33

    and for 4 < t < 6, s = -6.67

    and so the total DISTANCE is

    d = 66.7 + 1.33 + 6.67 = 74.7  ◄

    (meters, presumably)

    Note that if you do a single integration from -3 to 6 you get -72 (meters, presumably). That's the DISPLACEMENT during the time interval.

    If you find this helpful, please award Best Answer!

  • Anonymous
    1 month ago

    72 m....................

    • Andrew Smith
      Lv 7
      1 month agoReport

      This is the absolute value of the DISPLACEMENT, but it isn't the distance moved.

    • Login to reply the answers
  • Anonymous
    1 month ago

    total distance is ≅ 183.5 m

    • Andrew Smith
      Lv 7
      1 month agoReport

      No it isn't.  You might like to give a reason why you chose this value.

    • Login to reply the answers
  • 1 month ago

    Because they asked for DISTANCE ( not displacement) you must determine the times ( if any) at which it changes direction.  ( 2 and 4 seconds) now integrate. s(t) = -t^3/3 + 3t^2 -8t (+c)

    Find how far it moves in each of the three intervals  and add the absolute values of each of these motions.

    • Login to reply the answers
Still have questions? Get your answers by asking now.