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Homework Help: Instantaneous Velocity and Acceleration

  1. Aug 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
    a. The average velocity of the car from t=0s to t=10s
    b. The instantaneous velocity of the car at t=10s
    c. The average acceleration of the car at t=10s
    d. The instantaneous acceleration of the car at t=10.s

    2. Relevant equations

    3. The attempt at a solution
    in a.) I integrated the equation to get the position function
    then I substitute t = 10 s
    I got 120 m/s but the answer should be -72 m/s <- So I need help here

    b.) I derive the x(t) to return the original equation then I substitute t = 10 s right?
    c.) I'll simply derive the original equation (7.20m/s^2)+2(-0.720)(10) = -7.2 m/s^2
    d.) Use the original equation then plug in t = 10 s ? <- Help
  2. jcsd
  3. Aug 18, 2015 #2


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    If x(t) is the position of the car, why are the units m/s? That's your first problem.

    Exactly which question does -72 m/s answer?
  4. Aug 18, 2015 #3
    Where did the -72 m/s come from? I calculated that the distance covered over 10 sec was 120 m. So the average velocity should be 12 m/s. During the 10 sec, the instantaneous velocity is never zero, so the average certainly shouldn't be negative.

  5. Aug 18, 2015 #4
    I borrowed my friend's notebook because I studying about kinematic then I saw this example and resolved it. The answer on a letter A is -72(on his notebook) but I got 120 soooooo I posted in this forum to confirm which is the right answer
  6. Aug 18, 2015 #5
    oooops I meant to say 120 m
  7. Aug 18, 2015 #6
    So, from what you know now, was his answer right or not?
  8. Aug 18, 2015 #7
    First I was frustrated that I can't get -72 m which consuming all my time so I decided to post my silly question (sorry guys I have to do it time is gold right?)

    but I conclude that maybe he miscalculate his problem or something
    but it is confirm the answer in a.) is 120 m? I always get 120 m. Sooooooo I hope I'm right. This problem is already wasted 1 hour in my review session.
  9. Aug 18, 2015 #8
    What is the definition of average velocity? Does velocity have units of meters?

  10. Aug 18, 2015 #9
    I meant to say when you're getting x(t) the units should m or meters right? So I answered 120 m. Then I use average velocity = ΔX/ΔT soooo 120 m / 10 s = 12 m/s right?
    This is for letter A
  11. Aug 18, 2015 #10

  12. Aug 18, 2015 #11
    for letter B finding the in instantaneous velocity. I don't need to derive it because it is already in v(t)
    I'll just plug t = 10 s into v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2

    I got 0 m/s hmmm I find it odd to get 0 in instantaneous velocity
  13. Aug 18, 2015 #12
    Have you ever heard of speeding up, then slowing down, and finally stopping.

  14. Aug 18, 2015 #13
    Oh okay that make sense to prove that it I can get 0 m/s thank you!!!
  15. Aug 18, 2015 #14
    Very good job. You were very close on a.
    ..The integration was correct.
    ..Just a minor math error

    1) Remember, you are calculating average velocity for the 10 second period. Since the initial velocity is 0 m/s. The final velocity is 0 m/s. And all velocities in between are greater than zero there is no way the answer can be a negative number.

    2) You have to dispose of the C by applying the equation when t=0. Naturally C= 0 so you can dispense with it.

    3) You calculated x(t) at 10 seconds to be 120 Meters (not m/s). So you got the distance traveled in ten seconds correct.

    4) You are going for average velocity. so that is the distance traveled divided by the time. v avg = [x(10s) - x (0s) ] / [t(10s) - t(0s)]
    V avg = 120 m / 10 s = 12 m/s

    Question: Now how do you check the answer?
  16. Aug 18, 2015 #15
    What is the difference between AVERAGE acceleration and INSTANTANEOUS acceleration.
    {Perhaps c. should be from t=0s to t=10s}

  17. Aug 18, 2015 #16
    Practicing Latex

    [tex] v(t)=(7.20m/s^2) \times t-(0.720m/s^3) \times t^2[/tex]

    Distance traveled is the area under the curve.
    So he needed to integrate the velocity equation, which he did correctly.
    [tex] \int v(t) \, dt = x(t) = \frac {7.20m/s^2 \times t^2} {2} \quad + \quad \frac {-0.720m/s^3 \times t^3} {3} \quad+ \quad C [/tex]

    And Acceleration is the slope of the velocity line.
    So he needed to derive the velocity equation, which he did correctly.
    [tex] \frac {\partial v(t)} {\partial t}= a(t) = 7.20m/s^2 \quad + \quad 2 \times (-0.720)m/s^3 \times t \quad [/tex]
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