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Math Help - A few antiderivative application problems

  1. #1
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    A few antiderivative application problems

    Problem 1
    A car starts from rest at time t = 0 and accelerates at -0.6t + 4 meters/sec2 for 0<t<12. How long does it take for the car to go 100 meters?

    here is my work:
     a = -0.6t + 4
     v = -0.3t^2 + 4t + c
     v = -0.3t^2 + 4t
     s = -0.1t^3 + 2t^2 + K
     s = -0.1t^3 + 2t^2
     100 = -0.1t^3 + 2t^2
     t = 10 seconds

    Problem 2
    A 727 jet needs to attain a speed of 200 mph to take off. If it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be? (Assume constant acceleration and give your final answer in feet)

    here is my work:
    200 miles per hour =  293 \frac{1}{3} ft/sec

     293 \frac{1}{3} ft/sec divided by 30 sec =  9 \frac{7}{9} ft/sec  ^2

     a = 9 \frac{7}{9} ft/sec ^2
     v = 9 \frac{7}{9} t + c ft/sec
     v = 9 \frac{7}{9} t ft/sec
     s = 4 \frac{8}{9} t^2 + K
     s = 4 \frac{8}{9} t^2
     s(30) = 4 \frac{8}{9} * (30)^2
     s(30) = 4400 ft.

    Please check the conversions and overall answer.

    Problem 3
    On the moon, the acceleration due to gravity is about 1.6 m/sec ^2 (compared to  g \approx 9.8 m/sec ^2 on earth). If you drop a rock on the moon (with initial velocity 0), find formulas for:

    (a) Its velocity, v(t), at time t.
    (b) The distance, s(t), it falls in time t.

    here is my work:
     a = -1.6 m/sec ^2
     v = -1.6t + c m/sec
    (a)  v(t) = -1.6t m/sec
    (b)  s(t) = -0.8t^2 + K , where K is the initial height.

    Should I keep the constant K in this answer? and how about the constants in all the other problems? I'm confused whether I need them or not.. In the previous ones I assumed the initial values were 0. Can I do this?
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  2. #2
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    Quote Originally Posted by Jacobpm64 View Post
    Problem 1
    A car starts from rest at time t = 0 and accelerates at -0.6t + 4 meters/sec2 for 0<t<12. How long does it take for the car to go 100 meters?
    a=-.6t+4
    v=\int -.6t+4dt=.3t^2+4t+C
    At t=0 we have v=0 (meaning of "at rest").
    Thus, C=0.
    Thus,
    v=-.3t^2+4t
    s=\int -.3t^2+4t dt
    s=-.1t^3+2t^2+C
    Is the equation for the distance function.
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  3. #3
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    All right,

    But then how would you get a value for C?
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  4. #4
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    Quote Originally Posted by Jacobpm64 View Post
    All right,

    But then how would you get a value for C?
    Look at what he wrote he
    tell you that C=0, which leaves you with:

    s=-0.1\, t^3 +2\,t^2

    To find how long it takes to go 100 metres you set s=100 in the above equation and solve for t. Which gives you the
    following cubic to solve:

    -0.1\, t^3 +2\,t^2-100=0,

    which the observant will note has a root at t=10, and the other roota are either negative ( and so irrelevant) or at a time greater than 12 (and so also irrelevant).

    So it takes 10 seconds for the car to go 100 metres.

    RonL
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