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Find its position at the instants when velocity is zero

  1. Sep 8, 2009 #1
    1.The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6.

    x1 = ? (m)
    x2 = ? (m)


    2. v(t) = 9.6t - 0.6t^5



    3. I tried setting v(t) = 0 and got -2, 0, 2. None of these turned out to be the right answer...I'm not sure what I'm doing wrong here.
     
  2. jcsd
  3. Sep 9, 2009 #2
    If you take a look at the position formula x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6, you can see that the position x depends on time t.
    So, you have to find that position when velocity is equal to zero. That means that you have to find the moment t when velocity is equal to 0. Your velocity formula v(t) = 9.6t - 0.6t^5 is ok. Now you have to find moment t when velocity is 0.
    v(t) = 9.6t - 0.6t^5
    0=9.6t - 0.6t^5=t(9.6-0.6t^4)
    I hope this helps.
     
  4. Sep 9, 2009 #3
    Yeah, I figured that out this morning. I forgot to take the t out when I set the equation equal to zero...that was my big problem

    But now I have to find its acceleration at the instants when the car has zero velocity

    a(t) = 9.6 - 3t^4
    But I'm not really sure what to do here. I cant take out a t, so how am I suppose to get two values of t? And when I get the values, what equation do I put them in: x(t), v(t), or a(t)?
     
  5. Sep 10, 2009 #4
    When car has a zero velocity? Aren't those the same t's you have already calculated for x?
     
  6. Sep 10, 2009 #5

    drizzle

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    Gold Member

    zero velocity→ zero acceleration
    thus, a(t)=0=... ⇒t=? ⇒ x(t=?)=??
     
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