December 9th, 2016, 08:08 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  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 
December 9th, 2016, 09:13 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,623 Thanks: 1303 
sometimes it helps to sketch a velocity vs time graph ... (see attached) acceleration to max velocity takes three times longer than the slowdown back to v = 0 let $3t$ = speed up time; $t$ = slowdown time $s = \dfrac{a}{2}(3t)^2 + \dfrac{3a}{2}t^2 = 6at^2$ $\bar{v} = \dfrac{s}{4t} = \sqrt{\dfrac{s}{2}} \implies s = 8t^2$ $6at^2 = 8t^2 \implies a = \dfrac{4}{3} \, m/s^2$ 
December 9th, 2016, 01:06 PM  #3 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
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 01:09 PM. Reason: personal 
December 9th, 2016, 01:10 PM  #4 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
i understand it now thanks v much spent hours trying to work it out

December 9th, 2016, 01:13 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,623 Thanks: 1303 
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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