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Answer 1

Question: A train starting from rest and move with a uniform acceleration of 0.2 metre per second sq. for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Provided that:

• Acceleration = 0.2 m/s sq.
• Time = 5 minutes
• Initial velocity = 0 m/s

Don't be confused! Initial velocity cames as zero because the train starts from rest.

To calculate:

• Final velocity
• Distance travelled

Solution:

• Final velocity = 60 m/s
• Distance travelled = 9000 m

Knowledge required:

• SI unit of acceleration = m/s²
• SI unit of distance = m
• SI unit of time = seconds
• SI unit of velocity = m/s

Using concepts:

• Formula to convert min-sec
• First equation of motion
• Third equation of motion

Using formulas:

• 1 minute = 60 seconds

• 1st eqⁿ of motion → v = u + at

• 3rd eqⁿ of motion → v² - u² = 2as

Where, a denotes acceleration, u denotes initial velocity, v denotes final velocity, s denotes displacement or distance or height, t denotes time taken.

Required solution:

~ Firstly let us convert min-sec!

→ 1 minute = 60 seconds

→ 5 minute = 5 × 60 seconds

→ 5 minute = 300 seconds

• Henceforth, converted!

~ Now by using first equation of motion let us calculate the final velocity!

→ v = u + at

→ v = 0 + (0.2)(300)

→ v = 0 + 60

→ v = 60 m/s

→ Final velocity = 60 m/s

• Henceforth, done!

~ Now by using third equation of motion let us calculate the distance travelled!

→ v² - u² = 2as

→ (60)² - (0)² = 2(0.2)(s)

→ 3600 - 0 = 0.4s

→ 3600 = 0.4s

→ 3600/0.4 = s

→ 36000/4 = s

→ 9000 = s

→ s = 9000 m

→ Distance = 9000 metres

• Henceforth, done! $\:$