Question

A train has running 108 km is boat held in 2 minutes calculate the rate retardation produced by application of the brakes

Answer 1

Proper question: A train has running with the velocity of 108 km/h for 2 minutes. Calculate the rate of retardation produced by application of the brakes.

Provided that:

• Initial velocity = 108 km/h
• Time taken = 2 minutes
• Final velocity = 0 m/s

To calculate:

• The retardation

Solution:

• The retardation = -0.25 m/s²

Using concepts:

• Acceleration formula
• Formula to convert kmph - mph
• Formula to convert min into s

Using formulas:

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

• ${\small{\underline{\boxed{\sf{1 \: minute \: = 60 \: seconds}}}}}$

Required solution:

~ Firstly let us convert km/h into m/s

$:\implies \sf 108 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{108} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 6 \times 5 \\ \\ :\implies \sf 30 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let's convert min into seconds!

$:\implies \sf 1 \: minute \: = 60 \: seconds \\ \\ :\implies \sf 2 \: minute \: = 60 \times 2 \: seconds \\ \\ :\implies \sf 2 \: minute \: = 120 \: seconds \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

• Time taken = 120 seconds
• Initial velocity = 30 m/s

~ Now let's find retardation!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-30}{120} \\ \\ :\implies \sf a \: = \dfrac{-30}{120} \\ \\ :\implies \sf a \: = \dfrac{-3{\cancel{0}}}{12{\cancel{0}}} \\ \\ :\implies \sf a \: = \dfrac{-3}{12} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-3}{12}} \\ \\ :\implies \sf a \: = \dfrac{-1}{4} \\ \\ :\implies \sf a \: = -0.25 ms^{-2} \\ \\ :\implies \sf Acceleration \: = -0.25 ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.25 ms^{-2}$