Question

d) A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s2 . Find how far the car will go before it is brought to rest?
(i) 8100 m (ii) 900 m (iii) 625 m (iv) 620 m​

Answer 1

Answer:

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Answer 2

Solution

Given :-

• Speed of a car = 90 km/h.
• Acceleration = - 0.5 m/s².

Find :-

• how far the car will go before it is brought to rest?

Explanation

We Know,

• 1 km = 1000 m
• 1 hours = 3600 second.

So,

==> 90 km/h = 90 × 5/18 m/sec.

==> 90 km/h = 25 m/sec.

Using Formula

★ v = u + at.

In last movement, train will be in rest position.

so,

• v = 0 m/s².

Now, keep all values in formula,

==> 0 = 25 - 0.5t

==> 0.5 t² = 25

==> t = 25/0.5

==> t = 25×10/5

==> t = 50 sec.

Again using Formula

★ S = ut + at²/2

Where,

• s be distance of car till to stop .

So, keep all required values.

==> s = 25×50 + (-0.5)×50²/2

==> s = 1250 - 1250.0/2

==> s = 1250 - 625

==> s = 625 m

Hence

• Distance travel by car is (s) = 625 m
• Taken time be (t) = 50sec.

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