A spaceship is traveling directly away from earth with speed 4t2+6 km/s. At time t=0 is is 250 km from earth. How far is it from earth one m

Question

A spaceship is traveling directly away from earth with speed 4t2+6 km/s. At time t=0 is is 250 km from earth. How far is it from earth one minute after time t=0? Include units in your answer.

in progress 0
Jezebel 4 years 2021-08-05T09:24:28+00:00 2 Answers 28 views 0

Answers ( )

  1. taiduc
    0
    2021-08-05T09:26:10+00:00
    Reply

    Answer:

    The spaceship is 288,610km from the earth one minute after t = 0.

    L(60) = 288,610 km

    Explanation:

    Let L represent how far the spaceship is from the earth.

    The speed the spaceship can be given as;

    dL/dt = 4t^2 + 6

    So,

    L = ∫ dL/dt = ∫ 4t^2 + 6

    L(t) = (4t^3)/3 + 6t + C

    At t = 0, L(0) = 250 km

    L(0) = (4(0)^3)/3 + 6(0) + C = 250

    C = 250

    Then,

    L(t) = (4t^3)/3 + 6t + 250 …….1

    Since dL/dt is in m/s

    time t is in seconds.

    One minute after t = 0 is t = 1 minutes = 60 seconds.

    Substituting t = 60 into equation 1;

    L(60) = (4(60)^3)/3 + 6(60) + 250

    L(60) = (4×216000)/3 + 360 + 250

    L(60) = 288,610 km

    Therefore, the spaceship is 288,610km from the earth one minute after t = 0

  2. yenoanh
    0
    2021-08-05T09:26:13+00:00
    Reply

    Answer: 576610 km

    Explanation:

    the speed is dx/dt which is 4t² + 6

    On integration, we have,

    x = 8t³/3 + 6t + C

    C is the constant of integration,

    t is the time in second,

    x is the distance in km

    So, at t = 0, and x = 250, we have

    250 = 8(0)³/3 + 6 * 0 + C

    250 = 0 + 0 + C, so that,

    C = 250

    x = 8t³/3 + 6t + C, where C = 250

    x = 8t³/3 + 6t + 250

    at time, t = 1 min = 60 s

    x = 8(60³)/3 + 6(60) + 250

    x = (8*216000)/3 + 360 + 250

    x = 1728000/3 + 610

    x = 576000 + 610

    x = 576610 km

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )