The train leaving the station at an acceleration of 0.2 m / s2 reached

1.	The train leaving the station at an acceleration of 0.2 m / s2 reached a speed of 20 m / s. How far had the train come by then?
Answer» Given : • Acceleration of the train = 0.2 m/s² • INITIAL velocity of the train = 0 m/s • Final velocity of the train = 20 m/s To find : • Distance travelled by the train Solution : Here we SHALL use the third equation of motion to find the distance travelled by the train. Third equation of motion :- v² - u² = 2as where, • v denotes the final velocity • u denotes the initial velocity • a denotes the acceleration • s denotes the distance/displacement we have, • v = 20 m/s • u = 0 m/s • a = 0.2 m/s² Substituting the given values :- → (20)² - (0)² = 2(0.2)(s) → 400 = 2 × 2/10 × s → 400 = 4/10 × s → 400 × 10/4 = s → 100 × 10 = s → 1000 = s → The value of s = 1000 Therefore, the distance travelled = 1000 m

The train leaving the station at an acceleration of 0.2 m / s2 reached a speed of 20 m / s. How far had the train come by then?

Answer»

Given :

• Acceleration of the train = 0.2 m/s²

• INITIAL velocity of the train = 0 m/s

• Final velocity of the train = 20 m/s

To find :

• Distance travelled by the train

Solution :

Here we SHALL use the third equation of motion to find the distance travelled by the train.

Third equation of motion :-

v² - u² = 2as

where,

• v denotes the final velocity

• u denotes the initial velocity

• a denotes the acceleration

• s denotes the distance/displacement

we have,

• v = 20 m/s

• u = 0 m/s

• a = 0.2 m/s²

Substituting the given values :-

→ (20)² - (0)² = 2(0.2)(s)

→ 400 = 2 × 2/10 × s

→ 400 = 4/10 × s

→ 400 × 10/4 = s

→ 100 × 10 = s

→ 1000 = s

→ The value of s = 1000

Therefore, the distance travelled = 1000 m

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