Calc

The position of an object moving along a line is given by the function s left parenthesis t right parenthesis equals negative 4.9 t squared plus 31 t plus 15. Find the average velocity of the object over the following intervals.

​(a) left bracket 0 comma 6 right bracket
​(b) left bracket 0 comma 5 right bracket
​(c) left bracket 0 comma 4 right bracket
​(d) left bracket 0 comma 1 plus h right bracket​, where hgreater thanminus1 is any real number
​(a) The average velocity of the object over the interval left bracket 0 comma 6 right bracket is ________.

How do you find average velocity given the interval?

  1. 👍
  2. 👎
  3. 👁
  1. average velocity = displacement / time
    by the way NOT distance/time
    distance is scalar
    displacement is vector, might be - part of the time.

    if on earth and
    original height = 15 meters
    original speed up = 31 m/s and v = 31 - 9.8 t
    acceleration = -g = -9.8 m/s^2
    then
    h(t) = 15 + 31 t - (9.8/2)t^2
    agreed :)
    so here
    h(t) = 15 + 31 t -4.9 t^2
    1 to 6 seconds
    h(6) = 15 +186 - 176.4 =24.6
    h(0) = 15
    so average v =1.6 m/s
    onward - do the rest.
    by the way, when does it get to the top?
    v = 31 - 9.81 t
    v = 0 at top
    so t = 3.16 seconds then starts down
    If they had asked for SPEED, not Velocity, you would have had to do up from t = 0 to t = 3.16 then down from t = 3.16 to t = 6 and call both the positive and negative distances +

    1. 👍
    2. 👎
  2. The reason I went into that detail is I am not sure your question is worded correctly. I wonder if they really want the average speed, not the average velocity.

    1. 👍
    2. 👎
  3. if I go 4 miles from here to downtown at 30 miles/hour, then immediately come back at 30, my average velocity is ZERO. I went nowhere :)
    However my average SPEED was 30

    1. 👍
    2. 👎
  4. Great. Thank you for specifying the details.

    I do have another question. How can I find the average velocity over the interval [0, 1+h] given the function -4.9t^2 +31t+15. I am confused how to set this up and how would you do it by the graphing calculator.

    1. 👍
    2. 👎
  5. at 0 it is 15

    at (1+h) it is 15 + 31(1+h) - 4.9(1+h)^2
    subtract
    d moved = 31(1+h) - 4.9(1+h)^2
    divide by time (1+h)
    d moved/time = 31 - 4.9(1+h)

    1. 👍
    2. 👎
  6. It would have made much more sense to ask you
    what is average velocity from t = t to t = t+h ?
    s(t+h) = 15 + 31 (t+h) -4.9 (t^2+2 t h+h^2)
    s(t) = 15 + 31 t - 4.9 t^2
    then distance moved is
    s(t+h) - s(t) = 31 h - 9.8 t h - 4.9 h^2
    divide by time elapsed h
    [ s(t+h)-s(t) ]/h = 31 - 9.8 t - 4.9 h

    NOW what if I let h ---> 0 ???
    That is velocity at t , the definition of the derivative of displacement with time
    v = 31 - 9.8 t
    (which I automatically used a while ago )
    Surely that is what they want you to know.

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. physics

    The position versus time for a certain object moving along the x-axis is shown. The object’s initial position is −2 m. Find the instantaneous velocity at 6 s. Answer in units of m/s

  2. Calc 1

    Consider the position function s(t)=sin((pi)(t)) representing the position of an object moving along a line on the end of a spring. Sketch a graph of s with the secant line passing through (0,s(0)) and (0.5,s(0.5)). Determine the

  3. physics

    The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.3m/s3, B = 1.8m/s2, C = −4.3m/s, and D = 2 m.(a) What is the velocity of the object at t = 12s? (b) What is the

  4. physics

    An object moves along a straight line and the velocity as a function of time is presented by the above graph. a. Find the acceleration of the object for the time intervals: 0-5 s, 5 s- 10 s, 10 s-15 s, 15 s – 20 s, 20 s – 25

  1. Physics

    An object moving with uniform acceleration has a velocity of 13.0 cm/s in the positive x-direction when its x-coordinate is 2.73 cm. If its x-coordinate 2.95 s later is −5.00 cm, what is its acceleration? cm/s^2 Express the

  2. Calc!

    Direction: The table gives the position​ s(t) of an object moving along a line at time​ t, over a​ two-second interval. Find the average velocity of the object over the following intervals. a.​ [0, 2] b.​ [0, 1.5] c.​

  3. Calculus

    The position of a particle moving on a horizontal line is given by s(t)=2t^3-15t^2+24t-5, where s is measured in feet and t in seconds. a: What is the initial position of the particle? b: What is the average velocity of the

  4. physics

    In this problem you will determine the average velocity of a moving object from the graph of its position as a function of time . A traveling object might move at different speeds and in different directions during an interval of

  1. Physics

    Please show how you got the answers, not just the answers. A 40.0 N force stretches a vertical spring 0.250 m. (A) What mass must be suspended from the spring so that the system will oscillate with a period of 1.0 s? (B) If the

  2. chabot college

    Consider the position function s(t)=sin((pi)(t)) representing the position of an object moving along a line on the end of a spring. Sketch a graph of s with the secant line passing through (0,s(0)) and (0.5,s(0.5)). Determine the

  3. AP Calculus

    A particle is moving along a horizontal straight line. The graph of the position function (the distance to the right of a fixed point as a function of time) is shown below. Answer the following questions only on the interval

  4. physics

    The position versus time for a certain object moving along the x-axis is shown. The object’s initial position is −3 m. Find the instantaneous velocity at 8 s. Answer in units of m/s.

You can view more similar questions or ask a new question.