Question

A train is moving at 108 km / hr .after applying brake the speed of train reduce to 72 km/hr . within 20 second. find acceleration and distance ​

Answer 1

Answer:

Explanation:

Given,

Initial speed of the train, u = 108 km/h = 108 × 5/18 = 30 m/s

Final speed of the train, v = 72 km/h = 72 × 5/18 = 20 m/s

Time taken by the train, t = 20 seconds

To Find,

Acceleration by the train and,

Distance covered by the train.

Formula to be used,

1st equation of motion i.e, v = u + at

3rd equation of motion i.e, v² - u² = 2as

Solution,

Putting all the values, we get

v = u + at

⇒ 20 = 30 + a × 20

⇒ 20 - 30 = 20a

⇒ - 10 = 20a

⇒ - 10/20 = a

⇒ a = - 0.5 m/s²

Hence, the acceleration is - 0.5 m/s².

Now, the distance covered,

Putting all the values, we get

v² - u² = 2as

⇒ (20)² - (30)² = 2 × (- 0.5) × s

⇒ 400 - 900 = - 1s

⇒ - 500/- 1 = s

⇒ s = 500 m.

Hence, the distance covered is 500 m.

Answer 2

Answer:

$\huge \bf \: Given$

• Initial velocity (U) = 108 km/hr
• Final velocity (V) = 72 km/hr
• Time (T) = 20 Sec = 0.03 h

$\huge \bf \: To \: find$

❶ Acceleration of train

❷ Distance travelled by train

$\huge \bf \: Solution$

108 km/h = 108 × 5/18 = 30 m/s

72 km/h = 72 × 5/18 = 20 m/s

Now,

Finding acceleration

v = u + at

20 = 30 + a × 20

20 - 30 = 20a

-10 = 2 a

- 10/20 = a

a = - 0.5 m/s²

Now,

Finding

v² - u² = 2as

(20)² - (30)² = 2 × (- 0.5) × s

400 - 900 = - 1s

- 500/- 1 = s

s = 500 m.