Question

brakes are applied to train travelling at 72 km/h after passing over 200m . its velocity is reduced to 36km/h . find the retardation

Answer 1

Explanation:

Initial speed:- 72000m/3600s=20m/s

Final Speed:-36000m/3600s=10m/s

Distance travelled:- 200m

Let retardation=r

therefore, 100=400-400r(v^2=u^2+2as)

300=400r

r=3/4 m/s^2

Final speed =0

Initial speed= 10m/s

r=3/4m/s^2

therefore, using v^2=u^2+2as

0=100-3/2s

3/2s=100

s=100*2/3

s=66.6666667m

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Answer 2

Answer:

0.75m/s^2

Explanation:

V2=u2*2as

U in m/s=72*5/18=20m/s

V in m/s=36*5/18=10m/s

10^2=20^2+2*a*200

100=400+400a

100-400=400a

a=-300/400

Acceleration =-0.75m/s2

Retardation =0.75

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