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
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
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