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Math Help - acceleration/speed question

  1. #1
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    acceleration/speed question

    The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2
    for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree?

    I thought I understood acceleration and speed until this problem!

    Thanks!
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  2. #2
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    \Delta x = v_f t - \frac{1}{2}at^2

    solve for v_f
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \Delta x = v_f t - \frac{1}{2}at^2

    solve for v_f
    Thanks! Though I would be very grateful if you could say how you got there. I'm not sure what the variables are. I would love a few sentences!!
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  4. #4
    Super Member 11rdc11's Avatar
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    \Delta x = \text{final position - intital position}

    v_f = \text{final velocity}

    t = \text{time}

    a = \text{acceleration}
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    \Delta x = \text{final position - intital position}

    v_f = \text{final velocity}

    t = \text{time}

    a = \text{acceleration}
    Thanks! Is there a more general formula from which this formula is derived? Where does the 1/2 come from? Is it the formula for average acceleration?
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  6. #6
    Super Member 11rdc11's Avatar
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    I think from this postion function

    x(t) = \frac{at^2}{2} + vt + x

    which when you take the derivative is

    v(t) = at +v

    and then

    a(t) = a
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  7. #7
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    Quote Originally Posted by 2clients View Post
    The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2
    for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree?

    I thought I understood acceleration and speed until this problem!

    Thanks!
      {\text{Now, }}\overrightarrow a  = {\text{acceleration}} =  - 5.6{\text{ m/s}}^2  \hfill \\

      \overrightarrow {v_1 }  = {\text{ initial velocity of car}} \hfill \\

      \overrightarrow {v_2 }  = {\text{ final velocity of car}} \hfill \\

      \Delta t = {\text{ time interval }} = 4.2{\text{ s}} \hfill \\

      \Delta \overrightarrow d  = {\text{ displacement of car }} = 62.4{\text{ m}}{\text{.}} \hfill \\

      {\text{Now, using the formula, Acceleration, }}\overrightarrow a  = \frac{{\overrightarrow {v_2 }  - \overrightarrow {v_1 } }}<br />
{{\Delta t}} \hfill \\

      \overrightarrow {v_2 }  - \overrightarrow {v_1 }  =  - 5.6 \times 4.20 \hfill \\

      \overrightarrow {v_2 }  = \overrightarrow {v_1 }  - 23.52{\text{  }}......................{\text{eqn}}{\text{. (1)}} \hfill \\

      {\text{Now, Displacement, }}\Delta \overrightarrow d  = \overrightarrow {v_1 } \Delta t + \frac{1}<br />
{2}\overrightarrow a \left( {\Delta t} \right)^2  \hfill \\

      62.4 = \overrightarrow {v_1 } \left( {4.2} \right) + \frac{1}<br />
{2}\left( { - 5.6} \right)\left( {4.2} \right)^2  \hfill \\

      \overrightarrow {v_1 }  = 26.62{\text{  m/s}} \hfill \\

      {\text{Put this value in eqn}}{\text{.(1)}} \hfill \\

      \overrightarrow {v_2 }  = 26.62 - 23.52 \hfill \\

      \overrightarrow {v_2 }  = 3.1{\text{ m/s}} \hfill \\

     {\text{So, the speed of the car is 3}}{\text{.1 m/s, when it hits the tree}}{\text{.}} \hfill \\
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  8. #8
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    Quote Originally Posted by Shyam View Post
     {\text{Now, }}\overrightarrow a = {\text{acceleration}} = - 5.6{\text{ m/s}}^2 \hfill \\

     \overrightarrow {v_1 } = {\text{ initial velocity of car}} \hfill \\

     \overrightarrow {v_2 } = {\text{ final velocity of car}} \hfill \\

     \Delta t = {\text{ time interval }} = 4.2{\text{ s}} \hfill \\

     \Delta \overrightarrow d = {\text{ displacement of car }} = 62.4{\text{ m}}{\text{.}} \hfill \\

     {\text{Now, using the formula, Acceleration, }}\overrightarrow a = \frac{{\overrightarrow {v_2 } - \overrightarrow {v_1 } }}<br />
{{\Delta t}} \hfill \\

     \overrightarrow {v_2 } - \overrightarrow {v_1 } = - 5.6 \times 4.20 \hfill \\

     \overrightarrow {v_2 } = \overrightarrow {v_1 } - 23.52{\text{ }}......................{\text{eqn}}{\text{. (1)}} \hfill \\

     {\text{Now, Displacement, }}\Delta \overrightarrow d = \overrightarrow {v_1 } \Delta t + \frac{1}<br />
{2}\overrightarrow a \left( {\Delta t} \right)^2 \hfill \\

     62.4 = \overrightarrow {v_1 } \left( {4.2} \right) + \frac{1}<br />
{2}\left( { - 5.6} \right)\left( {4.2} \right)^2 \hfill \\

     \overrightarrow {v_1 } = 26.62{\text{ m/s}} \hfill \\

     {\text{Put this value in eqn}}{\text{.(1)}} \hfill \\

     \overrightarrow {v_2 } = 26.62 - 23.52 \hfill \\

     \overrightarrow {v_2 } = 3.1{\text{ m/s}} \hfill \\

     {\text{So, the speed of the car is 3}}{\text{.1 m/s, when it hits the tree}}{\text{.}} \hfill \\
    Thank you very much!!! It makes perfect sense now.
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  9. #9
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    derivation of \Delta x = v_f t - \frac{1}{2}at^2 ...

    \Delta x = v_{avg} t

    for constant a, v_{avg} = \frac{v_f + v_0}{2}

    \Delta x = \frac{v_f + v_0}{2}t

    \Delta x = \frac{v_f}{2}t + \frac{v_0}{2}t

    \Delta x = v_f t - \frac{v_f}{2}t + \frac{v_0}{2}t

    \Delta x = v_f t - \frac{v_f - v_0}{2}t

    \Delta x = v_f t - \frac{1}{2} \cdot \frac{v_f - v_0}{t}t^2

    \Delta x = v_f t - \frac{1}{2}at^2

    v_f = \frac{\Delta x}{t} + \frac{1}{2}at = \frac{62.4}{4.2} - (2.8)(4.2) = 3.1 \, m/s
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