Question

2-11
Two trains travelling on the same track are approaching each other with equal speeds of 40 m/s.
The drivers of the trains begin to decelerate simultaneously when just 2.0 km apart. Assuming
decelerations to be uniform and equal, the value of the deceleration to barely avoid collision
should be
(a) 0.8 m/s?
(b) 2.1 m/s? (c) 11.0 m/s?
(d) 11.8 m/s2

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Answer 1

Answer:

-0.8 m/sec^2

Step-by-step explanation:

initial velocity is 40m/s

to avoid collision on same track they need to completely stop,therefore final velocity will be zero.

distance is 2km i.e. 2000m.

using 3rd equation of motion

2as= v^2 - u^2

2*a*2000=-1600

a=-1600/2000

a=-0.4ms^-2

now total retardation for 2 trains will be into 2.

you get the answer

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Answer 2

** [in this case , both the trains are in similar motion ,but in opposite direction...

and so,they will have to travel the same distance while decelerating]

initial velocity of both trains=40 m/s (u)

final velocity for both trains will be =0 m/s (v)

Distance to be covered by each train is equal** and therefore=1000 m (s)

We have to find the retardation (given that it is equal in both trains):

      so using $v^{2} =u^{2} + 2as$...

             0  = 1600 + 2 (a) 1000

             (a) = $\frac{-1600}{2000}=-\frac{4}{5}$

Hence,acceleration=0.8 m/s for each train