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Math Help - time and distance ..please help with steps.

  1. #1
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    time and distance ..please help with steps.

    If A and B are running in a circular track in a direction opposite to that in which C is running, who is running at twice and thrice the speeds of A and B respectively and on the same track. They start running from a same point. It is known that A's average speed is 3 m/s and the track is 120 m.
    When after the start, will B find himself equdistant and between A and C for the first time?
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  2. #2
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    Re: time and distance ..please help with steps.

    Given the speed of A, find the speed of B and C.

    The circular track has 2*pi radians and 120 meters long so people running round it change angle at a rate of 2*pi/120= pi/60 radians per meter.
    If B is running with speed v then he is changing direction at a rate of v*(pi/60) radians per second.
    After a time t B will have gone v*t*(pi/60) radians
    Putting the circle in the x-y plane B's x co-ordinate is r*cos(v*t*(pi/60)) where r is the radius of the circle
    B's y co-ordinate will be r*sin(v*t*(pi/60))

    Likewise if A is running at speed u then his x and y co-ordinates will be r*cos(u*t*(pi/60)) and r*sin(u*t*(pi/60))

    C is running in the opposite direction at a speed w. Because its going the opposite direction you have to put a minus sign in front, so C's x co-ordinate is -r*cos(w*t*(pi/60)) and y co-ordinate is -r*sin(w*t*(pi/60))

    Now you have their co-ordinates in terms of time get an expression for the distance between their points.
    Distance |BC|= |AB|
    That gives you one equation in terms of time, solve the equation for t.

    Remember all angles are in radians.
    Thanks from hisajesh
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  3. #3
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    Re: time and distance ..please help with steps.

    Is there a way to solve by just using algebra?
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  4. #4
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    Re: time and distance ..please help with steps.

    time and distance ..please help with steps.-circle-1.pngtime and distance ..please help with steps.-circle-2.png
    I am sure you would understand the algebraic method and also notice how making a drawing makes things easy to understand.
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  5. #5
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    Re: time and distance ..please help with steps.

    Hello, hisajesh!

    A\text{ and }B\text{ are running in a circular track in a direction opposite to that in which }C\text{ is running,}
    \text{who is running at twice and thrice the speeds of }A\text{ and }B\text{, respectively, and on the same track.}
    \text{They start running from the same point at the same time.}
    \text{It is known that }A[\text{'s average speed is 3 m/s and the track is 120 m long.}
    \text{When after the start, will }B\text{ find himself equdistant and between }A\text{ and }C\text{ for the first time?}

    We find that: . \begin{Bmatrix}\text{A's speed}&=& \text{3 m/s} \\ \text{B's speed} &=& \text{2 m/s} \\ \text{C's speed}&=& \text{6 m/s} \end{Bmatrix}

    Picture the circular track; circumference 120 meters.
    The one-meter marks are labeled clockwise: 1, 2, 3 . . . ending with 120 at the top.
    A and B are running clockwise; C is running counter-clockwise, starting at the top.

    In t seconds, A has run to point 3t.
    In t seconds, B has run to point 2t.
    In t seconds, C has run to point 120-6t.

    When is B exactly halfway between A and C?
    When is B's coordinate the average of A's coordinate and C's coordinate?

    We have: . 2t \:=\:\frac{3t + (120-6t)}{2} \quad\Rightarrow\quad 4t \:=\:120-3t \quad\Rightarrow\quad 7t \,=\,120

    Therefore: . t \:=\:\frac{120}{7} \:=\:17\tfrac{1}{7}\text{ seconds.}
    Thanks from hisajesh
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