Question

A train starting from rest attains a velocity of 72 km h^-1 in 5 minutes. Assuming that the acceleration is uniform, find (1) the acceleration and (2) the distance travelled by for attaining this velocity.​

Answer 1

Given:-

• Initial velocity, u = 0m/s

• Final velocity , v = 72km/h

• Time taken, t = 5min

To Find:-

• (1) Acceleration , a

• (2) Distance travelled , s

Solution:-

Firstly we convert the unit here

Final velocity = 72×5/18 = 20m/s

Time taken = 5×60 = 300s

[1]

The rate of change of velocity at per unit time is called acceleration

→ a = v-u/t

Substitute the value we get

→ a = 20-0/300

→ a = 20/300

→ a = 1/15 m/s² or 0.06m/s²

∴ The acceleration of the train is 0.06m/s²

[2]

By using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 20² = 0² + 2×1/15 × s

→ 400 = 0 + 2/15 ×s

→ s = 400×15/2

→ s = 200×15

→ s = 3000m or 3km

∴ The distance covered by the train is 3km.

Answer 2

Answer:

3km or 3000m

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