Question

A train starting from rest attains a velocity of 72km/hr in 5 mins. Assuming that the acceleration is uniform. find:-
1) the acceleration and
2) the distance travelled by the train for attaining this velocity​

Answer 1

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✤ Required Answer:

✒ GiveN:

• The body started from rest.
• Attains a velocity of 72 km/hr
• Time taken = 5 mins

✒ To FinD:

• The acceleration of the body?
• Distance travelled by the train$?$

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✤ How to solve?

The above question is based on equations of motion which contains several measures in kinematics like Distance, Initial velocity, Final velocity, Accleration and Time. The three equations of motion are:

• v = u + at
• s = ut + 1/2 at²
• v² = u² + 2as

Here, a is acceleration, v is final velocity, u is initial velocity, s is the displacement/distance and t is the time taken by the body.

⛈ So, let's solve this question...

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✤ Solution:

We have,

• Initial velocity = 0 m/s (Rest)
• Final velocity = 72 km/hr

[ Unit conversion km/hr ➝ m/s ]

➝ 72 km/hr = 72 × 5/18 m/s

➝ 72 km/hr = 20 m/s

• Time taken = 5 mins = 300 seconds

By using 1st equation of motion,

➝ v = u + at

➝ 20 = 0 + a × 300

➝ 300a = 20

➝ a = 20/300 = 0.067 m/s²

✒ Acceleration of the body = 0.067 m/s²

Now we have,

• Accleration = 0.067 m/s²
• Initial velocity = 0 m/s
• Final velocity = 20 m/s

By using 3rd question of motion,

➝ v² = u² + 2as

➝ 20² = 0² + 2 × 20/300 × s

➝ 400 = 40s/300

➝ s = 400 × 300 / 40

➝ s = 3000 m

Or

We also have,

• Time taken = 300 s

By using 2nd equation of motion,

➝ s = ut + 1/2 at²

➝ s = 0 × 300 + 1/2 × 20/300 × 300 × 300

➝ s = 10 × 300

➝ s = 3000 m

✒ Distance travelled by the body = 3000 m

☀️ Henc, solved !!

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