Question

A train starting from rest attains a velocity of 72 kilometers per hour in 5 ,minutes ,Assuming that the acceleration is uniform .find the acceleration and find the distance travlled by the train for attaining this velocity

Answer 1

Given:-

• Initial velocity (u) = 0m/s

• Final velocity (v) = 72×5/18 = 20m/s

• Time taken (t) = 5×60 = 300s

To Find:-

• Acceleration (a) and Distance (s)

Solution :-

Firstly we calculate the acceleration of the train

By using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 20 = 0 + a×300

→ 20 = a×300

→ a = 20/300

→ a = 2/30 m/s² or 0.066m/s²

∴ The acceleration of the train is 0.066m/s².

Now Calculating the distance .So by using 3rd equation

→ v² = u²+2as

Substitute the value we get

→ 20² = 0² + 2×2/30 × 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 :-

• Acceleration = 0.066m/s²

• Distance = 3km

Given :-

• Initial velocity, u = 0

• Final velocity, v = 72 × 5/18 = 20m/s

• Time, t = 5 min = 5 × 60 = 300s

To Find :-

• Acceleration, a

• Distance, s

Solution :-

★ We know that

a = v - u / t

➪ a = 20 - 0/ 300

➪ a = 20/300

➪ a = 2/30

➪ a = 0.066m/s²

→ Acceleration of the train is 0.066m/s².

Let's calculate distance

v² = u² + 2as

➪ 20² = 0² + 2×2/30×s

➪ 400 = 0 + 2/15s

➪ s = 400 × 15/2

➪ s = 200 × 15

➪ s = 3000m or 3km

→ Distance of the train is 3000m or 3km.

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