Question

a bus is traveling at 72km/hr. brakes are applied and the bus comes to a stop after 10 minutes. find the retardation of the brakes

Answer 1

Answer:

Provided that:

• Final velocity = 0 mps
• Initial velocity = 72 kmph
• Time = 10 minutes

To calculate:

• The acceleration

Solution:

• The retardation = -0.03 m/s²

Using concepts:

• Formula to convert kmph-mps.

• Formula to convert min-sec!

• To calculate the acceleration we can use either first equation of motion or formula to calculate the acceleration.

Using formulas:

• Acceleration formula:

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

• First equation of motion:

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

• Formula to convert min-sec:

• ${\small{\underline{\boxed{\pmb{\sf{1 \: min \: = 60 \: sec}}}}}}$

~ Firstly let us convert kmph-mps!

$:\implies \sf 1 \: kmph \: = \dfrac{5}{18} \: mps \\ \\ :\implies \sf 72 \: kmph \: = 72 \times \dfrac{5}{18} \: mps \\ \\ :\implies \sf 72 \: kmph \: = \cancel{72} \times \dfrac{5}{\cancel{{18}}} \: mps \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf 72 \: kmph \: = 4 \times 5 \\ \\ :\implies \sf 72 \: kmph \: = 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Henceforth,

• Initial velocity = 20 mps
• Final velocity = 0 mps

~ Now let's convert min-sec!

$:\implies \sf 1 \: min \: = 60 \: sec \\ \\ :\implies \sf 10 \: min \: = 60 \times 10 \: sec \\ \\ :\implies \sf 10 \: min \: = 600 \: sec \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Henceforth,

• Time = 600 seconds

~ Let's calculate the acceleration now!

By using acceleration formula.

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-20}{600} \\ \\ :\implies \sf a \: = \dfrac{-20}{600} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-20}{600}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = \cancel{\dfrac{-2}{60}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = \dfrac{-1}{30} \\ \\ :\implies \sf a \: = -0.03 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -0.03 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.03 \: ms^{-2}$

By using first equation of motion.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 20 + a(600) \\ \\ :\implies \sf 0 - 20 = 10a \\ \\ :\implies \sf - 20 = 600a \\ \\ :\implies \sf a \: = \dfrac{-20}{600} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-20}{600}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = \cancel{\dfrac{-2}{60}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = \dfrac{-1}{30} \\ \\ :\implies \sf a \: = -0.03 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -0.03 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.03 \: ms^{-2}$