Question

a train is travelling at the speed of 180 km per hour .Brakes are applied so as to produce a uniform acceleration of 1.0 metre per second square .Find how far the train will go before it is bought to rest
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Answer 1

16200 m

initial velocity (u) = 180kmph (180*5/18)= 50m/s

acceleration(a) = -1mps२

final velocity (v) = 0 m/s

we know

${v}^{2} - {u}^{2} = 2as$

$0 - {50}^{2} = 2 \times ( - 1) \times s$

s = 16200m

or 16.2km

Answer 2

Answer:

Distance covered by the train, s = 1250 meters

Explanation:

It is given that,

Initial speed of the train, u = 180 km/h = 50 m/s

Brakes are applied, final speed, v = 0

Uniform acceleration, $a=1\ m/s^2$

As brakes are applied then it will produce a deceleration, $a=-1\ m/s^2$

We need to find the distance covered by the train before it is brought to rest. Let the distance covered be x. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

$0-(50\ m/s)^2=2\times -1\ m/s^2\times s$

s = 1250 m

So, the distance covered by the train before it comes to rest is 1250 meters. Hence, this is the required solution.