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Math Help - speed - distance catching up

  1. #1
    Senior Member DivideBy0's Avatar
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    speed - distance catching up

    A speeding motorbike travels past a stationary police car. The police car starts accelerating immediately, and keeps accelerating until it has passed the bike.

    DATA: motorbike speed: 35ms^{-1}; police car acceleration: 4ms^{-2}

    How far does the police car travel before it overtakes the motorbike?


    I tried working this out using the following method, but it is wrong. How come?

    My Solution

    Let the distance travelled by the motorbike be d_m and the distance travelled by the police car be d_p. Let t be time in seconds.

    The distance travelled by the motorbike is given by

    d_m=35t

    The distance travelled by the police car can be expressed by

    d_p=4+8+...+t=\sum_{i=1}^t 4i = \frac{4t(t+1)}{2}=2t(t+1)

    When d_p=d_m, 35t=2t(t+1)

    Solving gives t=16.5 \Rightarrow d = 577.5 \mbox{m}

    Thankyou!
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  2. #2
    Super Member

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    Hello, DivideBy0!

    Are we allowed to use Calculus?


    A speeding motorbike travels past a stationary police car. The police car
    starts accelerating immediately, and keeps accelerating until it has passed the bike.

    Data: motorbike speed: 35 m/s; police car acceleration: 4 m/s

    How far does the police car travel before it overtakes the motorbike?

    We have: . v_m \:=\:35
    . . Integrating: . d_m \:=\:35t + C_1
    At t = 0,\:d = 0\quad\Rightarrow\quad C_1 = 0
    . . Hence: . \boxed{d_m \:=\:35t}


    We have: . a_p \:=\:4
    . . Integrate: . v_p \:=\:4t + C_2
    At t=0,\:v = 0\quad\Rightarrow\quad C_2 = 0
    . . Hence: . v_p \:=\:4t

    Integrate: . d_p \:=\:2t^2 + C_3
    At t = 0,\:d = 0\quad\Rightarrow\quad C_3 = 0
    .Hence: . \boxed{d_p \:=\:2t^2}


    When are these two distances equal?
    . . 2t^2 \:=\:35t\quad\Rightarrow\quad 2t^2-35t \:=\:0\quad\Rightarrow\quad t(2t-35) \:=\:0

    . . Hence: . t \;=\;0,\:17.5

    The police car overtakes the motorbike in 17 seconds.


    The police car travels the same distance as the motorbike.

    . . d_p \;=\;d_m \;=\;35(17.5) \;=\;\boxed{612.5 \text{ m}}

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