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Related to Constant acceleration !

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose a child driving a go-cart is traveling 4m/s when she crosses a line 4m from her starting point. She continues with a steady acceleration of 0.4m/s^2 , until she crosses a mark 40m from the starting point. How long does it take for her to go from the 4m mark to the 40m mark?


    what i did is this :

    a = 0.4m/s^2
    Vi = 4m/s
    d = 40m - 4m = 36m

    i used this formula :


    Vf = Vi + a*t
    0 = 4m/s + (0.4m/s^2)t
    -0.4m/s^2 = 4m/s
    t = -10s

    and this :

    d = Vi+Vt/2 * t
    36m = (4m/s + 0)/2 * t
    36m = 2m/s t

    divide both side by 2m/s

    t = 18s


    but the true answer is 6.7s

    anyone correct this ?
     
  2. jcsd
  3. Apr 26, 2012 #2
    the direct way to solve it, without having to calculate the final velocity, is to use
    d = V1 t + a t2 /2
    put in the values d= 36, a=0.4 and solve the quadratic equation in t using the normal formula. Its obviously the positive value of t you take.

    Hope this helps

    Regards

    Sam
     
    Last edited: Apr 26, 2012
  4. Apr 26, 2012 #3
    how about the velocity ? Vi = 0 or 4m/s ?
     
  5. Apr 26, 2012 #4
  6. Apr 26, 2012 #5
    36=4t+0.2t^2
     
  7. Apr 26, 2012 #6
    just a confirmation ..

    correct this if i'm wrong ..

    d = 36
    a = 0.4

    i used :

    d = V1 t + a t2 /2
    subst:

    36 = 0 t + 0.4 t2 /2
    divide both sides by 0.4

    90 = t2 /2
    divide both sides by 1/2

    45 = t2
    square root both sides to remain t :

    t = 3√5 or 6.7

    is it correct usage of algebra ?
     
  8. Apr 26, 2012 #7
    try 36 = 4t+0.2t2

    0.2 t2 +4t -36 =0

    t = (-4 + √(16+4*0.2*36))/(2*0.2)

    t≈6.7 seconds
     
    Last edited: Apr 26, 2012
  9. Apr 26, 2012 #8
    whow ! again .. i should study back again .. so thanks Sam ..

    gratitude !
     
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