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Math Help - Finding the slope of this function(the slope is also avg. velcoity I think)

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    Finding the slope of this function(the slope is also avg. velcoity I think)

    A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by , where = 2.40 and = 0.100 .

    1. Calculate the average velocity of the car for the time interval = 0 to = 10.0 .

    2. How long after starting from rest is the car again at rest?

    I know this is borderline physics problem, but I need to find the slope of this function, which will give me the average velocity, but I do not really know how.

    When the time is 0, I received 2.3 meters.
    When the time is 10, I received 340 meters.
    However, when I did 337.7 meters divided by 10 seconds, I didn't get the right answer. Help is needed! thanks!
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by CrusaderKing1 View Post
    A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by , where = 2.40 and = 0.100 .

    1. Calculate the average velocity of the car for the time interval = 0 to = 10.0 .

    2. How long after starting from rest is the car again at rest?

    I know this is borderline physics problem, but I need to find the slope of this function, which will give me the average velocity, but I do not really know how.

    When the time is 0, I received 2.3 meters.
    When the time is 10, I received 340 meters.
    However, when I did 337.7 meters divided by 10 seconds, I didn't get the right answer. Help is needed! thanks!
    1. how did you get x(0) = 2.3 ?

    2. find when \frac{dx}{dt} = 0
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by skeeter View Post
    1. how did you get x(0) = 2.3 ?

    2. find when \frac{dx}{dt} = 0
    So use the derivative of the function(which is 2bt+3ct^2) and plug in 5.00 s(a point besides 0) and 10.0 so I get a slope of 0.3 by (5,16.5) and (10,18)?
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by CrusaderKing1 View Post
    So use the derivative of the function(which is bt+3ct^2) and plug in 5.00 s(a point besides 0) and 10.0 so I get a slope of 0.3 by (5,16.5) and (10,18)?
    huh?

    v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}

    the derivative of the position function is instantaneous velocity ... you do not need a derivative to find average velocity.
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by skeeter View Post
    huh?

    v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}

    the derivative of the position function is instantaneous velocity ... you do not need a derivative to find average velocity.
    t2 = 10.0
    t1= 0.00

    x(t2)= 140

    140/10 =14 ?
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  6. #6
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by CrusaderKing1 View Post
    t2 = 10.0
    t1= 0.00

    x(t2)= 140

    140/10 =14 ?
    14 m/s ... that's it, Zeke.
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    Re: Finding the slope of this function(the slope is also avg. velcoity I think)

    Quote Originally Posted by skeeter View Post
    14 m/s ... that's it, Zeke.
    Thanks a billion man! Ya, I had 3 questions on instantaneous velocity and I got them all right fairly fast, but average velocity was difficult for me. If I have a question on the last problem, ill ask again, but im going to try to figure it out first.
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