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particle accelerating


markosheehan

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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.

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If an object moves with constant acceleration, a, then its speed at time t is [math]v= at+ v_0[/math] where [math]v_0[/math] is its speed at t= 0. The distance it covered in that time is [math]d= (a/2)t^2+ v_0t[/math]. Those are what you are referring to as the "uvast" equations.

 

Here, the train starts from rest (so [math]v_0= 0[/math]) then moves at acceleration a for some unspecified time [math]t_1[/math]. At the end of that time it will have speed at and will have gone a distance [math](a/2)t^2[/math]. It then decelerates at -3a for some unspecified time [math]t_2[/math]. At the end of that time t will have speed [math]-3at_2+ (at_1)= a(t_1- 3t_2)= 0[/math] (since it decelerates to rest). Unless a is 0 (in which case the train does not move at all) we must have [math]t_1= 3t_2[/math].

 

The train will have traveled an additional distance [math](1/2)at_2^2+ at_1t_2[/math]. The total distance it will have traveled is [math]s= (a/2)t_1^2+ (a/2)t_2^2+ at_1t_2= (a/2)(t_1^2+ t_2^2)+ at_1t_2[/math]. Since [math]t_1= 3t_2[/math], we can write that as [math]s= 2at_2^2+ 3at_2^2= 5at_2^2[/math]. The total time taken is [math]t_1+ t_2= 3t_2+ t_2= 4t_2[/math] so the average speed is [math]\frac{s}{t_1+ t_2}= \frac{5at_2^2}{4t_2}= \frac{5}{4}at_2= \frac{5}{2}at_2^2[/math]. Solve that for [math]t_2[/math] then use one of the previous equations to find a.

 

 

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You have an equation that relates distance, speed and acceleration. You need to apply that twice, since you have a total length traveled, so you have this for each leg of travel)

 

vf2-v02 = 2a(dot)s

 

You know the total distance traveled, so you can find the length of each leg of travel.

 

From that information, you should be able to apply the equation HallsofIvy posted to get the total time of travel, and from that, the average speed.

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