Question

A train that starts from reset attains a velocity of 72km/h in 5minutes (i) the acceleration and (ii) distance​

Answer 1

Explanation:

Distance = 72*5=360km

a= v-u/t= 72-0/5=14.4m/s

Answer 2

Final Velocity = displacement/time

= 72/5 km/h

= 14.4 km/h

Acceleration = (final velocity - initial velocity)/time

= 14.4/5= 2.88 km/hr

Distance= 72X5/60 km

= 6 kms.

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