Question

A train starting from rest attains a velocity of 20 m/s in 5 minutes. Assuming
the acceleration is uniform,

Answer 1

Appropriate Question:

A train starting from rest attains a velocity of 20m/s in 5 minutes. Assuming that the acceleration is uniform, find

i) Acceleration

ii) Distance travelled by the train while it attained the velocity

Answer:

Acceleration = 0.066 m/s²

Distance = 3000 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 [As it starts from rest]
• Final velocity (v) = 20 m/s
• Time taken (t) = 5 minutes

We are asked to calculate,

• Acceleration
• Distance travelled by the train while it attained the velocity

In order to calculate acceleration and distance travelled, firstly we need to convert time taken into its standard form, that is seconds.

↠Time = 5 minutes

1 minute consists of 60 seconds. Then,

↠ t = (5 × 60) s

↠ t = 300 s

Therefore, time taken is 300 seconds.

Now, let's find out the acceleration. By using the first equation of motion,

$\\ \longrightarrow \quad \boxed{\pmb{\sf { v = u + at}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

↠20 = 0 + 300a

↠20 – 0 = 300a

↠20 = 300a

↠20 ÷ 300 = a

↠1 ÷ 15 = a

↠0.066 m/s² = a

∴ Acceleration of the train is 0.066 m/s².

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Now,we have to calculate the distance travelled. By using the third equation of motion.

$\\ \longrightarrow \quad \boxed{\pmb{\sf { v^2 - u^2 = 2as}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

↠(20)² – (0)² = 2 × 1/15 × s

↠400 – 0 = (2 ÷ 15)s

↠400 × 15 = 2s

↠6000 = 2s

↠6000 ÷ 2 = s

↠3000 m = s

∴ Distance travelled is 3000 m.

Answer 2

Correct Question :-

• A train starting from rest attains a velocity of 20m/s in 5 minutes. Assuming that the acceleration is uniform, find  acceleration  and distance travelled by the train while it attained the velocity.

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Given :-

Initial velocity of train is 0 m/s

Final velocity of trains is 20 m/s

Time taken by train is 5 minutes = 300 s

To Find :-

Acceleration of Train

Distance travelled by train.

Solution :-

❍ To calculate acceleration we have to substitute the values in the first equation of motion ::

          v = u + at

Where,

• v is final velocity

• u is initial velocity

• a is acceleration

• t is time taken

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Finding acceleration :-

↦ 20 = 0 + [ 300 × a ]

↦ 20 = 0 + 300a

↦ 300a = 20 - 0

↦ 300a = 20

↦ a = 20/300

↦ a = 2/30

↦ a = 1/15 m/s²

• Henceforth, the acceleration of the train is 1/15 m/s²

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❍ To calculate the distance travelled by train we have to substitute the values in the third equation of motion ::

   v² - u² = 2as

Where,

• v denotes final velocity

• u denotes initial velocity

• a denotes acceleration

• s denotes distance travelled

Finding distance travelled :-

↦ [ 20 ]²  - [ 0 ]² = 2 × 1/15 × s

↦ 400 - 0 = 2 × 1/15 × s

↦ 400 = 2 × 1/15 × s

↦ 400 = 2s/15

↦ 2s = 15 × 400

↦ s = 15 × 400/2

↦ s = 15 × 200

↦ s = 3,000 m

• Henceforth, the distance travelled by the train is 3,000 m.

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