Question

A train is moving with a velocity of 90 km h-!. It is
brought to stop by applying the brakes which
produce a retardation of 0.5 m s? Find : (i) the
velocity after 10 s, and (ii) the time taken by the
train to come to rest. Ans. (i) 20 m s-1, (ii) 50 s​

Answer 1

Answer:

i) v = 0

a = - 0.5 m/s²

t = ?

u = 90 × 5/18 m/s = 25 m/s

v = u + at

0 = 25 + (- 0.5) t

0= 25 - 0.5t

0.5 t = 25

t = 25 / 0.5

t = 50 s

ii) u = 90 km/h = 90 × 5/18= 25 m/s

a = - 0.5 m /s²

t = 10 s

v = u + at

v = 25 + ( - 0.5) 10

v = 25 - 5 = 20 m/s

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Explanation:

Answer 2

i) Here, u = 90km/hr = 25m/s

a = -0.5m/s²

t = 10s

and, v = the velocity after 10 seconds

Considering, v = u+at

The velocity, v = 25+(-0.5×10)

= 25-5

= 20m/s

ii) Here, u = 90km/hr = 25m/s

a = -0.5m/s²

v = 0m/s

Considering, v = u+at

Time taken for the train to come to rest will

be:

0 = 25+(-0.5t)

0 = 25-0.5t

0.5t = 25

Therefore, t = 25/0.5

= 50s

Hope it helps....