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Homework Help: Motion in one dimension problem

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    An object moves along the x axis according to the equation x = 3.00t2 - 2.00t + 3.00 , where x is in meters and t is in seconds.
    Determine :
    (a) the average speed between t = 2.00 s and t = 3.00 s.
    (b) the instantaneous speed at t = 2.00
    (c) the average acceleration between t = 2.00 s and t = 3.00 s.
    (d) the instantaneous acceleration at t = 2.00 s
    (e) at what time is the object at rest?

    2. Relevant equations
    average speed = [itex]\frac{d(total distance travelled)}{\Delta t}[/itex]

    instantaneous speed = [itex]\frac{dx}{dt}[/itex]

    [itex]\overline{a}[/itex] (average acceleration) = [itex]\frac{\Delta v}{\Delta t}[/itex]

    a (instantaneous acceleration) = [itex]\frac{dv}{dt}[/itex] = [itex]\frac{dx^2}{dt}[/itex]

    3. The attempt at a solution

    (a) I couldn't calculate the total distance.

    (b) instantaneous speed = instantaneous velocity = [itex]\frac{dx}{dt}[/itex] = 6.00t - 2.00
    instantaneous speed at t=2.00 = 6.00(2)-2.00 = 10 m/s

    (c) [itex]\overline{a}[/itex] = [itex]\frac{\Delta v}{\Delta t}[/itex] = [itex]\frac{vf-vi}{1}[/itex] = [itex]\frac{v(3)-v(2)}{1}[/itex] = [itex]\frac{16-10}{1}[/itex] = 6 m/s2

    (d) a = [itex]\frac{dx^2}{dt}[/itex] = 6 => a at 2.00s = 6 m/s2

    (e) the rest means that v = 0
    => 6t-2=0
    =>t=0.33 s //it seems wrong.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 1, 2012 #2
    Looks like you did pretty well.

    The total distance? x= distance,
    so calculate x at 3 secs minus x at 2 secs
  4. Nov 1, 2012 #3
    but x at 3 minus x at 2 is the displacement which we use to calculate the average velocity not average speed. right??
  5. Nov 1, 2012 #4
    Speed is just (absolute) velocity.
  6. Nov 1, 2012 #5
    I think no.\
    the average velocity is different from average speed
    average velocity = dispacement/time
    average speed = distance/time
    sometimes average velocity is zero (when a particle moves and return to the starting point) where the average speed is a positive number.

    in this problem I think the distance is like what you said = x at 3s - x at 2s because the position x is always increasing (from the equation)
  7. Nov 1, 2012 #6
    Speed is defined as the scalar magnitude of the velocity vector. Since this is a one dimensional problem it simply becomes the absolute velocity.
  8. Nov 1, 2012 #7
    sorry, still not convinced!
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