calculus

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Two runners, A and B run on a straight race track for 0</=t</=10 seconds. The graph below, which consists of two line segments, shows the velocity, in meters per second, of runner A. The velocity of runner B, in meters per second, is given by the function v defined by v(t)=24t/2t+3/

Although I can't post the graph, there are two points labeled, (3,10) and (10,10).

ind the velocity of runner A and runner B at t=2 seconds. I know you can't see the graph, but is there any way you could guide me on how to do this problem?? thank you

  • calculus -

    sorry the slash after the three should actually be a period, the equation is only 24t/2t+3

  • calculus -

    v(t)=24t/(2t+3) [[ Maybe ?????? ]]

    Although I can't post the graph, there are two points labeled, (3,10) and (10,10).
    [[ and (0,0) ]] maybe????

  • calculus -

    MAKING THOSE ASSUMPTIONS ABOUT WHAT YOU WERE GIVEN!!!!
    At two seconds for runner A
    I think runner A went from (0,0) to (3,10) in a straight line
    then:
    y = m x + b
    (0,0) ---> b = 0
    m = 10/3
    so
    y = (10/3)x
    for x = 2 y = 20/3 or 6.67

    Runner B if y = 24 x / (2x+3)
    where x = 2
    y = 48 / (4+3) = 48/7 = 6.86

  • calculus -

    thank you so much! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

  • calculus-follow up for damon -

    thank you so much or your help! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

  • calculus -

    No!!!
    acceleration = dv/dt

  • calculus -

    dv/dt for A = 10/3 (constant slope until t = 3)

    dv/dt for B
    v(t)=24t/(2t+3)
    a = dv/dt = [(2t+3)(24) -24t(2) ]/(2t+3)^2
    put in t = 2
    a(2) = dv/dt = [ 7(24)- 96 ] /(49)
    = 72/49 = 1.47

  • calculus -

    By the way, position is the anti-derivative of velocity

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