Question

A train starting from rest moves with a uniform acceleration of 0.2
m/s^2 (s to the power 2) for 5 minutes.
Calculate the speed acquired.

Answer 1

Solution :—

Given :-

• Initial Velocity (u) = 0 m/s
• Acceleration (a) = 0.2 m/s²
• Time (t) = 5 min = 300 sec

To Find :-

• Final Velocity (v)

Formula Used :-

• v = u + at

Thus, we will substitute the Values

⟹ v = 0 + 0.2 × 300

⟹ v = 0 + 60

⟹ v = 60

Therefore, the final velocity is 60 m/s.

Answer 2

Answer -

Given -

• u = 0 ( starting from rest )
• a = 0.2 m/s²
• t = 5 min

where

u is initial velocity.

a is acceleration.

t is time taken.

To find -

Final velocity - v

Formula used -

1st equation of motion - v = u + at

Solution -

u = 0 ( starting from rest )

a = 0.2 m/s²

t = 5 min = 5 × 60 = 300 sec

Substituting the value in 1st equation of motion -

v = u + at

v = 0 + 0.2 × 300

v = 2 × 30

v = 60 m/s

Speed aquired by train after 5 min is 60 m/s.

━━━━━━━━━━━━

Additional information -

s = ut + 1/2 at²

v² = u² + 2as

Snth = u + a/2 ( 2n - 1 )

━━━━━━━━━━━━