IIT JEE , AIEEE Forum - Physics , Mechanics

zeeshanp

a train starting from rest accelerates unifromly for 100s,runs at a constant speed for 5 minutes and then comes to a stop with unifrom retardation in the next 150 seconds. during this motion it covers a distance of 4.25 km. find its constant speed, acceleration and deceleration.

pls solve completely

!!!

ashish_banga

Is the answers

constant speed = 10m/sec

acceleration = 1/10 MKS units

deceleration = 1 / 15

if correct then tell i will post the solution

pink_ele

let it attians speed v
den
total distance=dis...1+dis...2+dis....3
4250=1/2a10^4+100a1x300+100a1 x150-1/2 a2 (150)^2
eqn 1
also
(100 a1)=a2 x 150
solve dis n u'll get d ans

ashish_banga

let acceleration in first 100 s = a

then distance covered in first 100 s = 1/2 a X 100 X100
= 5000a m

velocity at the end of 100s = a X 100
= 100a m/ s

distance covered by it when it is moving with constant speed
= 100 a X 5 X 60 m
= 30000 m

let deceleration = A
then 0 = v - A X 150
v = A X 150
100 a = A X 150
A = 2/3 a
distance traveled during this time interval = 1/2 A X 150 X150
= 7500a

total distance traveled = 5000a + 30000a + 7500a
= 42500 a
= 4250 m
this gives a = 1/10 m/ second square

rest u can find

allamraju

Let the acceleration be a,velocity be v and deceleration be d,then

v=0+100a=100a and 0=v-150d $\Rightarrow$ v=150d=100a $\Rightarrow$ d=2a/3

Now,Total distance(S)=S₁+S₂+S₃ where S₁=a(100)²/2,S₂=300v=3X10⁴a,S₃=v(150)-d(150)²/2=(150)(100a)-a(150)²/3.Substitute S=4.25km=4250m and add S₁,S₂,S₃.We get,

4250=(42500)(a) $\Rightarrow$ a=1/10=0.1m/s²,V=10m/s,d=2/30m/s²

allamraju

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Let the acceleration be a,velocity be v and deceleration be d,then

v=0+100a=100a and 0=v-150d $\Rightarrow$ v=150d=100a $\Rightarrow$ d=2a/3

Now,Total distance(S)=S₁+S₂+S₃ where S₁=a(100)²/2,S₂=300v=3X10⁴a,S₃=v(150)-d(150)²/2=(150)(100a)-a(150)²/3.Substitute S=4.25km=4250m and add S₁,S₂,S₃.We get,

4250=(42500)(a) $\Rightarrow$ a=1/10=0.1m/s²,V=10m/s,d=2/30m/s²

Oh...I didn't notice the above answers.So,now I notice that I need to increase my typing speed.

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Let the acceleration be a,velocity be v and deceleration be d,then

v=0+100a=100a and 0=v-150dv=150d=100ad=2a/3

Now,Total distance(S)=S1+S2+S3 where S1=a(100)2/2,S2=300v=3X104a,S3=v(150)-d(150)2/2=(150)(100a)-a(150)2/3.Substitute S=4.25km=4250m and add S1,S2,S3.We get,

4250=(42500)(a)a=1/10=0.1m/s2,V=10m/s,d=2/30m/s2

Oh...I didn't notice the above answers.So,now I notice that I need to increase my typing speed.

v=0+100a=100a and 0=v-150d $\Rightarrow$ v=150d=100a $\Rightarrow$ d=2a/3

Now,Total distance(S)=S₁+S₂+S₃ where S₁=a(100)²/2,S₂=300v=3X10⁴a,S₃=v(150)-d(150)²/2=(150)(100a)-a(150)²/3.Substitute S=4.25km=4250m and add S₁,S₂,S₃.We get,

4250=(42500)(a) $\Rightarrow$ a=1/10=0.1m/s²,V=10m/s,d=2/30m/s²