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- Thread starter Gadget
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Thank You

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you're correct that under constant acceleration (a), the x position is given by:Gadget said:

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[tex] x \ = \ x_{0} + v_{0} \Delta t + \frac{a ( \Delta t )^{2}}{2} [/tex]

or:

[tex] \left ( x - x_{0} \right ) \ = \ v_{0} \Delta t + \frac{a \left ( \Delta t \right )^{2}}{2} [/tex]

where Δt is the elapsed time from when the car was at initial position x

let x

[tex] ( x_{1} - x_{0} ) \ = \ v_{0} ( t_{1} - t_{0} ) + \frac{a ( t_{1} - t_{0} )^{2}}{2} [/tex]

[tex] ( x_{2} - x_{0} ) \ = \ v_{0} ( t_{2} - t_{0} ) + \frac{a ( t_{2} - t_{0} )^{2}}{2} [/tex]

where:

(t

(x

(t

(x

SOLVE for (a) by placing these values in the above 2 equations and eliminating v

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- #3

VietDao29

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I wonder if you could read this from graph...geosonel said:... (x1 - x0) =Read From Graph...

Another hint if you like : You should assume at the time t_0 = 0, the car starts to go. That means at t_0 = 0s, v_0 = 0 m/s.

[tex]v = v_0 + at = at[/tex] as [tex]v_0 = 0 \mbox{m / s}[/tex]

so at t = 8 seconds [tex]v_8 = 8a[/tex].

Can you find the distance the car goes between t = 8 and t = 12 with respect to a? From there, you can easily solve the problem.

Hope you get it.

Viet Dao,

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- #4

mukundpa

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If you can get slop of the tangent at t = 8 second that will be velocity at that time.

- #5

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Gadget, i know your question has been answered but nevertheless i would like to point your attention to this intro on Newtonian Physics. Perhaps it might be of some interest to you in the near future .Gadget said:

Thank You

Newtonian Mechanics Intro

regards

marlon

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