Question

a train is travelling at a speed of 90 km hour brakes are applied so as to produce a uniform accelleration of 0.5 metre second find how far the train will go before it is brought to rest​

Answer 1

u=90kmph=90×1000/60×60=5mps

v=0mps

a=0.5mps²

Using 3rd eqn of motion,

2as=v²-u²

2×0.5×s=5²-0²

1×s=25-0

s=25m

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Answer 2

Given , Final velocity = 0

Initial Velocity = 90 km/hr = 90 × 5 / 18 = 25 m/s.

Acceleration =

$0.5m {s}^{ - 2}$

Using 3rd equation of motion

=>

${v}^{2} = {u}^{2} + 2as \\ {v}^{2} - {u}^{2} = 2as \\ \frac{ {v }^{2} - {u}^{2} }{2a} = s \\ \frac{ {0}^{2} - ({25}^{2}) }{2 \times 0.5} = s \\ \\ \frac{0 - 625}{ </em><em>-</em><em>1} = s \\ \\ </em><em>625m = s$