At a distance L=400m away from the signal light, brakes are applied to a locomotive moving with a velocity, u=54km/h. determine the position of rest of the locomotive relative to the signal light after 1 min of the application of the brakes if its acceleration a= -0.3 m/s2 ??
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Ignoring s=ut+1\2at2
u=54 km/h=15m/s
a=-0.3 m/s
t=60s
s=ut+1/2at2
=(15)(60) +0.5(-0.3)(60)(60)
=540+(-900)
=460m
But s from signal at start=400 m
Therefore s at rest=460-400=60m
Pls mark as brainliest :D
u=54 km/h=15m/s
a=-0.3 m/s
t=60s
s=ut+1/2at2
=(15)(60) +0.5(-0.3)(60)(60)
=540+(-900)
=460m
But s from signal at start=400 m
Therefore s at rest=460-400=60m
Pls mark as brainliest :D
pruthvi7:
can you explain how 15m/s comes
step by step
Look, first you find the total distance traveled whenb the brakes are applied. You'll find it to bre greater than the original distance from the signal post. Subtract the two, and you'll get tyour answer. Hope you understood
u=54km/h=54*5/18=15m/s please solve this step by step and send I can't understand easily help me please
What do I have to explain haha
why and how 5/18 comes ?
Actually, if you divide 1000 by 3600, you get 5/18(1000 and 3600 are used to convert km to m and hr to seconds, I hope you know that)
thank you chiefLucky
No problem
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