Question

A train moving at 54km/h is brought to rest in 1min on application of brake . Find the retardation. Find the distance travelled by the train after application of break( please say fast)
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Answer 1

retardation is negative acceleration

v = 0

u = 15

t = 60 seconds

$a \: = \frac{v - u}{t}$

$a = \frac{0 - 15}{60}$

$a = \frac{ - 15}{60}$

a = -0.25metre per second square

Answer 2

initial velocity , u = 54 km/hr = 15 m/s

final velocity, v = 0

time = 1 min = 60s

using v = u + at

retardation = - 15/60 = -0.25 m/s²

using s = ut + 1/2at² we get

s = 900 - 450 = 450 m