A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest.
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Answer u is 90 km so 90×5/18m
v is 0
a is -0.5m
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Solution
Given :-
- initial velocity of train (u) = 90 km/h
- Acceleration of train (a) = (-0.5) m/s²
- last velocity of train (v) = 0 m/s²
Find :-
- Distance , cover by train .
Explantion
We Know,
★ 1 km = 1000 m
★1 hour = 3600 seconds
Now, we convert km/h in m/sec.
So,
==> 1 km/h = 1000/3600 m/sec.
==> 1 km/h = 10/36 m/sec.
==> 1 km/h = 5/18 m/sec.
So, Now
==> 90 km/h = 90 × ( 5/18)
==> 90 km/h = 5 × 5
==> 90 km/h = 25 m/sec.
Now, Using Formula
★ v² = u² + 2as
Where,
- v = last velocity
- u = initial velocity
- a = acceleration
- s = Distance
Keep here, required values
==> 0² = (25)² + 2 × (-0.5) × s
==> 2 × (-0.5) × s = 625
==> 1.0 × s = 625
==> s = 625/1
==> s = 625 m
Hence
- The train will take rest after 625 m cover distance .
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Note :-
- Negative acceleration means , the object will going slowing down and after some time object take rest .
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