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Homework Help: Caculating time

  1. Sep 1, 2004 #1
    A Honda Civic travels in a straight line along a road. Its distance [tex]x[/tex]from a stop sign is given as a function of time [tex]t[/tex] by the equation x(t) = 1.53m/s^2*t^2 - 5.10×10-2 m/s^3*t^3

    Question #1. Calculate the average velocity of the car for the time interval [tex]t=0[/tex] to [tex]t1=1.93s[/tex]. answer is in m/s

    here's what i think i need to do:

    plug in 0 and 1.93 for t... ok so if i plug in 0 for t, it would be zero, so i skip that and go on to plug in t1=1.93s for the equation.

    x(t)= 1.53m/s^2(1.93)^2-5.10 X10^-2m/s^3(1.93s)^3
    5.69m/s^2 - 36.66 X 10^-2 m/s^3
    x(t)=5.3m/s^2 <--- can m/s^2 be substracted from m/s^3?

    ok so would that be the time? how would i caculate the average velocity as stated in question #1?

    i think i need to use the formula Vf(Final velocity) - V i(initial velocity)/time

    ok so x(t) would be the time right? how would i find the final velocity and the initial velocity
    Last edited: Sep 1, 2004
  2. jcsd
  3. Sep 1, 2004 #2


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    To find the average speed you need to know the speed of the vehicle. Hint: differentiate!
  4. Sep 1, 2004 #3

    x(t) = 1.53m/s^2*t^2 - 5.10×10-2 m/s^3*t^3
    ok so d/dt*x(t) =...

    v(t) = 1.53(2)m/s*2(t) - 5.10 X 10^-2 (3)m/s^2*(3)t^2

    is that correct? what am i suppose to do with v(t)?
  5. Sep 1, 2004 #4


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    Now find the average! Hint: Add up the values of the speed (aka integration!) and divide by the time interval!
  6. Sep 2, 2004 #5
    the time interval is x(t)=5.3m/s^2 right? did i even do that correctly?

    and for the values of the speed, what am i integrating extactly? it cant be v(t), cause that would get me x(t), right? i am so confused
  7. Sep 2, 2004 #6
    first of all average velocity equals d/t so once you figure out
    x(1.93) then divide it by 1.93 to get the average velocity.

    you made some errors in your calc:

    x(1.93)= 1.53m/s^2(1.93s)^2-5.10 X10^-2m/s^3(1.93s)^3

    the units seconds cancel out, so you're left with meters.
    5.69m - 36.66 X 10^-2 m

    x = 5.3m

    This is an easy question and doesn't require ne calculus, which i haven't taken yet.
    Last edited: Sep 2, 2004
  8. Sep 2, 2004 #7


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    That will work too but I guessed this was for a beginning calculus course. A more advanced student will recognize that [itex]\int \frac {dx}{dt} dt[/itex] is just [itex]\Delta x[/tex]. I guess you're a more advanced student! ;-)
  9. Sep 2, 2004 #8
    ah perfect... thanks alot, it was easily than i thought. but how did the s^2 and s^3 cancel out?
  10. Sep 2, 2004 #9

    Actually, I haven't done ne thing with calculus yet. I'm actually still in high school and i'm planning to take calculus this coming school year.
  11. Sep 2, 2004 #10

    x(1.93)= 1.53m/s^2(1.93s)^2-5.10 X10^-2m/s^3(1.93s)^3

    As you can see, in (1.53m*(1.93s)^2)/s^2 and (10^-2*(1.93s)^3)/s^3

    the units s^2/s^2 and s^3/s^3 = 1 so they can be ignored.
  12. Sep 2, 2004 #11


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    Oops! Sorry about that. I KNOW you'll enjoy calculus! Keep up the good work.
  13. Sep 2, 2004 #12


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    try [tex] A = \frac{1}{2gt^2}[/tex]Therefore [tex]t^2=\frac{1}{2gA}[/tex]
  14. Sep 2, 2004 #13
    x(0)=0 m
    x(1.93)=5.33 m
    Average velocity, [tex]\bar{v}=\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}[/tex]

    =2.76 [tex]m/s[/tex]
  15. Sep 3, 2004 #14


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    I am latex challenged and used the wrong expression to boot. :blushing:
    [tex]D = v_it+\frac{at^2}{2}[/tex]

    [tex]D =[/tex] distance, m
    [tex]v_i =[/tex] initial velocity, m/s
    [tex]t =[/tex] time, s
    [tex]a =[/tex] acceleration, m/s
  16. Sep 3, 2004 #15


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    [tex]{\rm Average\ velocity}\ v_{avg}=\frac{\int v\ dt}{\int dt} = \frac{\Delta x}{\Delta t}[/tex]
    (This has units of velocity, as expected.)
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