a train accelerate 36km/k to 72km/p on covering 50m . Calculate the acceleration of a train and time taken to cover the distance
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INITIAL VELOCITY = 36km/h
FINAL VELOCITY= 72km/h
DISTANCE= 50metres
ACCELERATION=?
TIME=?
To find acceleration use 3rd equation of motion
i.e.
v^2 = u^2+ 2as
Substitute the values given
72^2 =36^2 + 2×a×50
5184 = 1296 + 100a
5184-1296 =100a
3888/100 = a
a= 38.88m/s^2
for finding time use 1st equation of motion
v=u+at
72 = 36 +38.88t
72-36= 38.88t
36/38.88= t
t = 0.925925.....seconds
FINAL VELOCITY= 72km/h
DISTANCE= 50metres
ACCELERATION=?
TIME=?
To find acceleration use 3rd equation of motion
i.e.
v^2 = u^2+ 2as
Substitute the values given
72^2 =36^2 + 2×a×50
5184 = 1296 + 100a
5184-1296 =100a
3888/100 = a
a= 38.88m/s^2
for finding time use 1st equation of motion
v=u+at
72 = 36 +38.88t
72-36= 38.88t
36/38.88= t
t = 0.925925.....seconds
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