Question

a train is travelling at a speed of 60km/hr breaks are applied so as to produce a uniform acceration of -0.5m/s² find how far the train will go before.

Answer 1

Let's convert the kmh−1 in ms−1

90kmh−1=90⋅10003600=90036=25ms−1

We use the equation,

v2=u2+2as
where v=final velocity (0)
u= initial velocity (25ms−1)
a= acceleration (−0.5ms−2)
s= distance =?

So, 0=252+(2⋅−0.5⋅s)

s=252=625m