Question

A train travelling at 72 kmph deaccelerate uniformly at 2 m/s² how much time does it take to stop?​

Answer 1

Provided that:

• Initial velocity = 72 kmph
• Acceleration = -2 mps sq.
• Final velocity = 0 mps

Don't be confused! Acceleration cames in negative because the train retards or decelerates.

To calculate:

• The time taken

Solution:

• The time taken = 10 seconds

Using concept:

• First equation of motion
• Formula to convert kmph-mps

Using formula:

• ${\small{\underline{\boxed{\sf{v \: = \: u \: + at}}}}}$

• ${\small{\underline{\boxed{\sf{Kmph \: = \dfrac{5}{18} \: mps}}}}}$

Where, v denotes final velocity,u denotes initial velocity, a denotes acceleration and t denotes time taken

Required solution:

~ Firstly let us convert kmph-mps!

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

→ v = u + at

→ 0 = 20 + (-2)(t)

→ 0 = 20 + (-2t)

→ 0 = 20 -2t

→ 0 - 20 = -2t

→ -20 = -2t

→ 20 = 2t

→ 20/2 = t

→ 10 = t

→ t = 10 seconds

→ Time = 10 seconds