Physics, asked by 1111111gokul1111111, 2 months ago

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.

Answers

Answered by anshikahooda7
0

Answer u is 90 km so 90×5/18m

v is 0

a is -0.5m

Attachments:
Answered by BrainlyYuVa
5

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

= + 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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