train accelerating

markosheehan · 2016-12-09 00:38:22

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

thickhead · 2016-12-09 06:22:56

Average speed =v/2. Does it help?

abhishek1996 · 2016-12-09 10:44:36

a=2/3 is the answer!
Note: t1 t2 s1 s2 are the times and distance covered in acceleration and deceleration process.

v=0+at1 and 0=v-3at2 implies t1=3t2 also s/(t1+t2)=sqrt(s/2) which gives t1+t2=sqrt(2s) putting t1=3t2 we get 4t2=sqrt(2s) and t2^2=s/8

now, v^2=2as1 and 0=v^2-6as2 implies s1=3s2 also s1+s2=s putting s1=3s2 gives 4s2=s i.e s2=s/4

we had v=3at2(see 3rd line) so v^2 =9(a^2)(t2^2)=6as2(see 4th line) now putting t2^2=s/8(see 3rd line) and s2=s/4(see 4th line)hm in this eqn we get a=2/3

thickhead · 2016-12-09 13:34:19

Hi abhishek1996,
I get a=4/3.

Bob · 2016-12-09 20:12:10

hi markosheehan

If you use v = u + at for each part of the motion you can easily work out the time to the change of motion (t1) in terms of the total time t2.

The average speed is S/t2 which will give you an expression for (t2)^2 in terms of s.

Finally use s = u(t1) + 0.5a(t1)^2 for the first part and almost everything cancels out, leaving you with the value of 'a'. I agree with thickhead's answer.

Bob

iamaditya · 2016-12-25 22:55:15

Hi. I Also agree with thickhead's answer. I solved like this:

v1=v m/s
u1= 0m/s (Starting from rest)
a1=a m/s²

v²=u²+2as
⇒s=(v²-u²)/2a
⇒s1=(v1²-u1²)/2a1
⇒s1=v²/2a

v=u+at
⇒v1=u1+a1t1
⇒(v1-u1)/a1=t1
⇒v/a=t

v2=0 m/s (Stopping at rest)
u2= v m/s
a2= 3a m/s²

v²=u²+2as
⇒s=(v²-u²)/2a
⇒s2=(v2²-u2²)/2a2
⇒s2=(-u2²)/2a2
⇒s2=(-v²)/6a

v=u+at
⇒v2=u2+a2t2
⇒(v2-u2)/a2=t2
⇒-u2/a2=t2
⇒-v/3a=t2

Avg. Speed=Total distance/Total time
=(s1+s2)/(t1+t2)
=v/2

v/2=√(s1+s2)
⇒v/2=v/√(3a)
⇒2=√(3a)
⇒4=3a
⇒a=4/3 m/s²

Last edited by iamaditya (2016-12-25 22:56:02)

markosheehan · 2016-12-26 06:36:52

at the end you write v/2=√(s1+s2) ⇒v/2=v/√(3a) should it not be v/2=√(s1+s2)/√2 which would be v/2=√(2v^2/3a)/√2 when i solve this i dont get a=4/3

iamaditya · 2016-12-26 22:47:58

Hey, I'm very sorry it was my silly mistake . Abhishek is right. Correct answer is 2/3 ms-²

markosheehan · 2016-12-27 02:02:15

however at the back of the book it says the answer is 4/3 . im confused

zetafunc · 2016-12-27 04:10:26

Are you using an M2 textbook by any chance?

markosheehan · 2016-12-27 06:11:30

i dont know what m2 means

thickhead · 2016-12-29 22:35:11

iamaditya wrote:

Hi. I Also agree with thickhead's answer. I solved like this:

v2=0 m/s (Stopping at rest)
u2= v m/s
a2= 3a m/s²
v²=u²+2as
⇒s=(v²-u²)/2a
⇒s2=(v2²-u2²)/2a2
⇒s2=(-u2²)/2a2
⇒s2=(-v²)/6a
v=u+at
⇒v2=u2+a2t2
⇒(v2-u2)/a2=t2
⇒-u2/a2=t2
⇒-v/3a=t2
Avg. Speed=Total distance/Total time
=(s1+s2)/(t1+t2)
=v/2
v/2=√(s1+s2)
⇒v/2=v/√(3a)
⇒2=√(3a)
⇒4=3a
⇒a=4/3 m/s²

v2=0 m/s (Stopping at rest)
u2= v m/s
a2=- 3a m/s²

v²=u²+2as
⇒s=(v²-u²)/2a
⇒s2=(v2²-u2²)/2a2
⇒s2=(-u2²)/2a2
⇒s2=(v²)/6a
s1+s2=v²/2a+v²/6a=2v²/3a

Last edited by thickhead (2016-12-29 22:37:52)

Math Is Fun Forum

#1 2016-12-09 00:38:22

train accelerating

#2 2016-12-09 06:22:56

Re: train accelerating

#3 2016-12-09 10:44:36

Re: train accelerating

#4 2016-12-09 13:34:19

Re: train accelerating

#5 2016-12-09 20:12:10

Re: train accelerating

#6 2016-12-25 22:55:15

Re: train accelerating

#7 2016-12-26 06:36:52

Re: train accelerating

#8 2016-12-26 22:47:58

Re: train accelerating

#9 2016-12-27 02:02:15

Re: train accelerating

#10 2016-12-27 04:10:26

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#11 2016-12-27 06:11:30

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