A train is travelling at a speed of 60 km/h.brakes are applyed as to produce a uniform acceleration of -0.5 m/s2 . fond how far will the train go before it is brought to rest .
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Answered by
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U=60km/h=300/18m/s
V=0
A(retardation)=-0.5m/s2
We know,
2as=v2-u2
2×(-0.5)×s=0-(300/18×300/18)
-1.0×s=-277.77
S=277.77m
I tried hope this will help
*calculation is done verbally it might be wrong overlook it.
If this helped please mark this brainliest
V=0
A(retardation)=-0.5m/s2
We know,
2as=v2-u2
2×(-0.5)×s=0-(300/18×300/18)
-1.0×s=-277.77
S=277.77m
I tried hope this will help
*calculation is done verbally it might be wrong overlook it.
If this helped please mark this brainliest
Answered by
2
Given
⚠️ Initial speed of train = 60 km/h
⚠️ Acceleration = -0.5 m/s²
⚠️ It's brought to rest..
To find
⚠️ Distance covered by train ..
Solution
Now we have ;
⇒ Initial speed (u) = 60 km/h = (60 × 5/18) m/s = 16.67 m/s
⇒ Acceleration (a) = -0.5 m/s²
⇒ Final speed (v) = 0 m/s
Now using 3rd equation of motion :
✒️ v² - u² = 2as
✒️ 0² - (16.67)² = 2 × (-0.5)s
✒️ 0 - 277.89 = -s
✒️ -277.89 = -s
✒️ s = 277.89 m
Therefore, distance covered by train before coming to rest = 277.89 m
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