Physics, asked by mausamyadav55662, 11 months ago

A train starts from rest and accelerates uniformly for 20 second to attend a velocity of 90 km per hour it travels at constant velocity for 15 minutes . the driver applies brake to stop the train in next 5s . find the total displacement off the train

Answers

Answered by Sciencelover23
31

Answer:

22812.5m

Explanation:

Initial acceleration =25÷20=5/4m/s^2

Dist travelled (s)=ut+1/2at^2

=0+1/2×5/4×20×20=250m

Dist travelled when the body was travelling with constant velocity of 90km/hr or25m/s for 15min= 25×900= 22500m

deceleration of the body (v=u+at)v=0

Therefore u=-25=a×5

a=-25÷5=-5m/s^2

V^2-u^2=2as

0-25^2=2×-5×s

S=25×25÷5×2=62.5

Therefore total distance covered =22500+62.5+250=22812.5m

Answered by amritpandey4321
4

Answer:

22812.5

Explanation:

INITIAL ACCELERATION=25÷20=5/4m/s^2

Distance travelled when the body was travelling with constant velocity of 90km/hr or 25m/s for 15min=25×900=22500m

deceleration of the body

(v=u+at) v=0

Therefore. u=25=a×5

a=25÷5=-5m/s^2

v^2-u^2=2as

o-25^2=2×5×s

S=25×25÷5×2=62.5

Therefore total distance covered= 22500+62.5+250=22812.5 m

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