Question

A train starting from rest attains a velocity of 70km/hr in 5 min. Assuming that acceleration is uniform. Find the acceleration and distance traveled

Answer 1

Given that, a train starts from rest (means the initial velocity of the train is 0 m/s) and attains a velocity of 70 km/hr. Here, final velocity of the train is 70 km/hr.

Also given that, time taken by train is 5 min.

We have to find the acceleration and distance travelled by the train.

Firstly, convert the km/hr into m/s. To do so, multiply the given value by 5/18.

⇒ v = 70 × 5/18 = 19.44 m/s

To convert min into seconds. Multiply the given value by 60.

⇒ t = 5 × 60 = 300 sec

Using the First Equation Of Motion,

v = u + at

Substitute the known values in the above formula,

→ 19.44 = 0 + a(300)

→ 19.44/300 = a

→ 0.06 = a

Therefore, the acceleration of the train is 0.06 m/s².

Now, for distance:

Using the Second Equation Of Motion,

s = ut + 1/2 at²

Substitute the known values,

→ s = 0(300) + 1/2 × 0.06 × (300)²

→ s = 0 + 0.03 × 90000

→ s = 2700

Therefore, the distance covered by the train is 2700 m.

Answer 2

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GIVEN:

• Train starting from rest
• And attains a velocity of 70km/h
• Time taken 5min
• Acceleration is uniform

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TO FIND:

• Acceleration
• Distance traveled

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CONVERSION:

Converting 70km/h → m/s

To convert we have to multiply with 5/18

70 × 5/18 = 19.444

Converting 5min → seconds By

1 min = 60

5 min = 60 × 5

5min = 300s

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SOLUTION:

Given, the train starts from rest that means initial velocity = 0 m/s

Using the kinematic equation

v = u + at

Where

• v : final velocity
• u : initial velocity
• a : acceleration
• t : time taken

→ 19.44 = 0 + a(300)

→ 19.44 = 300a

→ a = 19.44/300

→ a = 0.064 m/s²

Now, finding distance travelled using the kinematic equation

s = ut + 1/2 at²

Where

• s : distance travelled
• u : initial velocity
• a : acceleration
• t : time taken

→ s = 0(300) + 1/2 (0.064)(300)²

→ s = 0.5 × 0.064 × 90000

→ s = 0.03 × 90000

→ s ≈ 2700

Hence, acceleration = 0.0648m/s² & distance travelled = 2700 m

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