Question

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 .

Answer 1

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.
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Answer 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