Question

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 1

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