A train is travelling at a speed of 60 kilometre per hour brakes are applied so as to produce a uniform acceleration of negative 0.5 metres per second square find how far the train will go before it is bought at rest.
Answers
Answered by
47
Given:
- Initial velocity of train, u = 60 km/h = 60 × 5/18 = 100/6 m/s
- Uniform acceleration produced on applying brakes, a = -0.5 m/s²
- Final velocity of train, v = 0 (since it is brought to rest)
To find:
- Distance travelled by train before coming to rest and after applying brakes, s =?
Formula required:
- Third equation of motion
2 a s = v² - u²
[ Where a = acceleration, s = distance covered, v = final velocity, u = initial velocity ]
Calculation:
Using third equation of motion
→ 2 a s = v² - u²
→ 2 ( -0.5 ) s = ( 0 )² - ( 100/6 )²
→ - s = - ( 100/6 )²
→ s = 10000 / 36
→ s = 277.78 m
Therefore,
- Train will cover a distance of 277.78 meters after applying brakes and before coming to rest.
Answered by
8
Answer:
By using formula: 2as=v²-u²
Putting the values :
2(0.5)s=(0)²-(100/6)²
s=-(100/6)²
s= 10000/36
s= 277.78m
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