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Calculating velocity with force

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A net horizontal force F=A+Bt^3 acts on a 2kg object, where A=5N and B =2N/s^3. What is the horizontal velocity of this object 4 seconds after it starts from rest?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    F=ma=5N+(2N/s^3)t^3
    (2kg)a=5N+(2N/s^3)(4s)^3 ==> a= 2.5N + t^3 = 66.5m/s^2

    a=dv/dt ==> v=(2.5 N)t + (1/4 N)t^4
    when t= 4s
    V= (2.5)(4) +(1/4)(4)^4 = 74m/s

    I used V=Vi+at when Vi=0m/s, t=4s, a=66.5m/s^2, and I got V=266m/s

    Which one is correct way of solving it, and why? Please let me know. Thanks
     
  2. jcsd
  3. Sep 30, 2015 #2

    RUber

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    Homework Helper

    It seems like the first method is right.
    V = Vi + at only applies when acceleration is constant .
     
  4. Sep 30, 2015 #3

    RUber

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    I mean, ## v(t) = v(0) + \int_0^t a(\tau) d\tau ##, which as you correctly pointed out is
    ## v(t) =0+ \int_0^t \frac{ F(\tau)}{m} d\tau \\
    \quad = \int_0^t 2.5 + \tau^3 d\tau \\
    \quad = \left. (2.5\tau + \frac14 \tau^4) \right|_0^t \\
    \quad =2.5t + \frac{t^4} {4} ##

    Since acceleration is increasing (quickly) with time, it should be pretty clear why if you use a(4) as your constant acceleration in the equation:
    v(t)=v_i+at,
    the result would be so much larger than you found with the correct method.
     
  5. Sep 30, 2015 #4
    Yeah you are right. I just assumed the acceleration was constant that's why the result was incorrect. Thanks guys!
     
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