Question

a train is travelling at a speed of 90 km/h break are applied so as to produce a uniform acceleration of 0.5 ms2 in find how far the train will go before its broad tourist​

Answer 1

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Explanation:Answer: u = 90 km/h --> 90 * 5/18 = 25 m/s

vc= 0 m/s

a = -0.5 m/s² (as the train is decelerating)

By using the 3rd equation of motion, v² - u² = 2as

Substituting the values,

0² - 25² = 2 * (-0.5) * s

0 - 625 = -1s

-625/-1 = s

so, s = 625m

 

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Answer 2

Answer:

625 m

Explanation:

Given, initial velocity (speed) is 90 km/hr, acceleration is -0.5 m/s² (as breaks are applied) and final velocity is 0 m/s.

To find: distance covered by the train.

Convert km/hr into m/s.

1 km/hr = 5/18 m/s

90 km/hr = 90 × 5/18 = 25 m/s

Using the third equation of motion,

v² - u² = 2as

Substitute the values,

→ (0)² - (25)² = 2(-0.5)s

→ 0 - 625 = -s

→ -625 = -s

→ 625 = s

Hence, the distance covered by the train is 625 m.