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Thread: uniform acceleration

  1. #1
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    uniform acceleration

    a train accelerates from rest to a speed v m/s. it continues at this constant speed for a certain time and then decelerates uniformly to rest. if the average speed for the whole journey is 5v/6, show that the 4/5 of the whole distance is covered at a constant speed.

    i cant work this out. i am trying to make use of the formulas average speed = total distance/total time and s=t(u+v)/2 . i am working these out with the letters but it aint working out
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  2. #2
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    Re: uniform acceleration

    Let the train start from rest at $t=0$.
    Let $t_1$ be the time the train reaches a speed of $v$.
    Let $t_2$ be the time the train starts to slow down.
    Let $t_3$ be the time comes to a stop.
    (see velocity graph attached)

    Two expressions for total distance ...

    (1) $d_T = \dfrac{5v}{6} \cdot t_3$

    (2) $d_T = \dfrac{vt_1}{2} + v(t_2-t_1) + \dfrac{v(t_3-t_2)}{2}$

    for (2), first term is distance traveled during speed up; second term is the distance traveled at the constant speed $v$,; third term id the distance traveled slowing down


    simplifying equation (2) yields $d_T = \dfrac{v}{2} \cdot t_3 + \dfrac{v}{2}(t_2-t_1)$

    set this expression equal to equation (1) ...

    $ \dfrac{v}{2} \cdot t_3 + \dfrac{v}{2}(t_2-t_1) = \dfrac{5v}{6} \cdot t_3$

    solving for $(t_2-t_1)$ yields $(t_2-t_1) = \dfrac{2t_3}{3}$


    distance traveled at a constant speed is $d_C = v(t_2-t_1) = v\cdot \dfrac{2t_3}{3} = \dfrac{2vt_3}{3}$


    now calculate the ratio $\dfrac{d_C}{d_T}$ ...


    edit: you can also get $d_T$ by just calculating the area of the trapezoid shown in the graph ...

    $d_T = \dfrac{v}{2}\left[t_3 + (t_2-t_1)\right]$ ... which is the same as equation (2); might be easier to deal with.
    Attached Thumbnails Attached Thumbnails uniform acceleration-motion_prob.jpg  
    Last edited by skeeter; Oct 7th 2016 at 10:44 AM.
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    Re: uniform acceleration

    i still do not understand how does finding out the ratio between the constant distance/the total time get you anywhere. i am trying to find out the ratio of the distance travelled at a constant all over the total distance and the multiplying this by the total time but i dont have an expression for total time i am really confused on this question
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    Re: uniform acceleration

    $d_C$ represents the distance the train travels at a constant speed

    $d_T$ represents the total distance the train travels, not time.

    $\dfrac{2vt_3}{3} \div \dfrac{5vt_3}{6} = \, ?$
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    Re: uniform acceleration

    it equals 4/5. thanks i understand now
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  6. #6
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    Re: uniform acceleration

    Two questions: Does "uniform acceleration" mean constant acceleration? And are we to assume that the train accelerates from 0 to the constant speed at the beginning at constant acceleration (the problem only states that the train "decelerates uniformly to rest").
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  7. #7
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    Re: uniform acceleration

    Yes, constant accelerations on the speed up & slow down ... typical physics kinematics problem at this level. If not, either a velocity or acceleration function during speed up would be required.
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