Question

the speed of a train is reduced from 60 km/hat the same time as it travels a distance of 450 m.if the retardation is uniform,find how much further it will travel before coming to rest?

Answer 1

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Given,
 • Initial Velocity = u₀ = 60 kmph = 16.6667 mps
 • Final Velocity = u₁ = 15 kmph = 4.1667mps
 • Distance = S = 450m

♦ Defining 'a' and 't' as the variables for retardation and time period for covering the '450m' distance and Using first and second equations of motion, we arrive at a trivial conclusion :

$a = \frac{4.1667 - 16.6667}{t} =\ \textgreater \ t = \frac{-12.5}{a} \\ \\ S = 60 ( -12.5/a ) + 1/2 a ( -12.5/a )^2 \\ \\ =\ \textgreater \ 450 = ( -750/a ) + ( 78.125 / a )$

Solving the above trivial equation, we approximate our 'a' to -1.5 m s⁻⁻²

Now, an application of the Third Equation for u₂ = 0 and a = -1.5 gives :
  ◘  2( -1.5 ) S = - ( 60 )²
=> S = 3600 / 3 = 1200 m

Hence, we conclude, the train further covers ( 1200 - 450 ) = 750m distance


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The above 0.01 error in the retardation occurs due to taking the Velocities as their Decimal Expansion.
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Hope that helps