Question

A train is travelling at a speed of 90 km per hour Breakers are applied in the pain so as to produce a uniform acceleration of 0.5 metre per second square find distance for the train before it is brought to rest

Answer 1

u = 25 m/s

v = 0 m/s

a = - 0.5 m/s

so, s = (v^2 - u^2)/2a

     s = (625)/1

    s = 625 metres

Answer 2

Answer:

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/s

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2 Using equation of motion: v2 = u2 + 2as

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2 Using equation of motion: v2 = u2 + 2asWe have: 0 = (25)^2 + 2x(-0.5)xs

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2 Using equation of motion: v2 = u2 + 2asWe have: 0 = (25)^2 + 2x(-0.5)xs s = 625 m

initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/sfinal speed, v = 0 m/s uniform acceleration, a = -0.5 m/s2 Using equation of motion: v2 = u2 + 2asWe have: 0 = (25)^2 + 2x(-0.5)xs s = 625 m [s: distance traveled by the train after the brakes are applied and before it is finally brought into rest]