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December 9th, 2016, 09:08 AM   #1
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particle accelerating

a train accelerates uniformly from rest to a speed v m/s with uniform acceleration a m/s^2 it then decelerates uniformly to rest with uniform retardation 3a m/s^2. the total distance travelled is s metres. if the average speed for the whole journey is square root (s/2) find the value of a. does anyone know how to work this out using uvast equations

i also used the forumula a/d = t2/t1 i got t2=t1*a/3 and t1=3t2/a and i got the max speed =3t2 cause v/a=3t2a
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December 9th, 2016, 10:13 AM   #2
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sometimes it helps to sketch a velocity vs time graph ... (see attached)

acceleration to max velocity takes three times longer than the slow-down back to v = 0

let $3t$ = speed up time; $t$ = slow-down time

$s = \dfrac{a}{2}(3t)^2 + \dfrac{3a}{2}t^2 = 6at^2$

$\bar{v} = \dfrac{s}{4t} = \sqrt{\dfrac{s}{2}} \implies s = 8t^2$

$6at^2 = 8t^2 \implies a = \dfrac{4}{3} \, m/s^2$
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File Type: jpg vel_time_graph.jpg (10.0 KB, 1 views)
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December 9th, 2016, 02:06 PM   #3
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can you explain to me how you calculated the distance travelled during the acceleration and the deacceleration

Last edited by markosheehan; December 9th, 2016 at 02:09 PM. Reason: personal
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December 9th, 2016, 02:10 PM   #4
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i understand it now thanks v much spent hours trying to work it out
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December 9th, 2016, 02:13 PM   #5
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during acceleration ...

$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$

$\Delta x_1 = 0 \cdot (3t) + \dfrac{1}{2}a(3t)^2 = \dfrac{9}{2}at^2$

during deceleration ...

$\Delta x = v_f \cdot t - \dfrac{1}{2}at^2$

$\Delta x_2 = 0 \cdot t - \dfrac{1}{2}(-3a)t^2 = \dfrac{3}{2}at^2$


$s = \Delta x_1 + \Delta x_2$
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