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**markosheehan****Member**- Registered: 2016-06-15
- Posts: 51

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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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,077

Average speed =v/2. Does it help?

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**abhishek1996****Member**- Registered: 2016-12-09
- Posts: 1

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

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,077

Hi abhishek1996,

I get a=4/3.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

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

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 659

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)*

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**markosheehan****Member**- Registered: 2016-06-15
- Posts: 51

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

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 659

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

Practice makes a man perfect.

There is no substitute to hard work

All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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**markosheehan****Member**- Registered: 2016-06-15
- Posts: 51

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

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Are you using an M2 textbook by any chance?

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**markosheehan****Member**- Registered: 2016-06-15
- Posts: 51

i dont know what m2 means

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,077

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/s²v²=u²+2as

⇒s=(v²-u²)/2a

⇒s2=(v2²-u2²)/2a2

⇒s2=(-u2²)/2a2

⇒s2=(-v²)/6av=u+at

⇒v2=u2+a2t2

⇒(v2-u2)/a2=t2

⇒-u2/a2=t2

⇒-v/3a=t2Avg. Speed=Total distance/Total time

=(s1+s2)/(t1+t2)

=v/2v/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)*

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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