1.

an object travels with speed of 72 km /hr applied brakes and it stop after 10 s find the acceleration of bus and distance travelled by the bus and distance travelled by the bus (please explain the answer also)​

Answer»

ANSWER:

Provided that:

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

To calculate:

  • The acceleration
  • The distance travelled

Solution:

  • The acceleration = -2 m/s²
  • The distance travelled = 100 m

Using concepts:

• Formula to CONVERT kmph-mps.

• To calculate the acceleration we can USE either first equation of MOTION or formula to calculate the acceleration.

  • Choice may vary!

• To calculate the distance we can use either third equation of motion or SECOND equation of motion.

  • Choice may vary!

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}}}}}}

• Second equation of motion:

  • {\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}

• Third equation of motion:

  • {\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 = 2as}}}}}}

• Formula to convert kmph-mps:

  • {\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}

Required solution:

~ 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

~ Let's calculate the acceleration first!

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}{10} \\ \\ :\implies \sf a \: = \dfrac{-20}{10} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-20}{10}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2}

By using first equation of motion.

:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 20 + a(10) \\ \\ :\implies \sf 0 - 20 = 10a \\ \\ :\implies \sf - 20 = 10a \\ \\ :\implies \sf a \: = \dfrac{-20}{10} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-20}{10}} \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2}

~ Let's calculate distance now!

By using third equation of motion.

:\implies \sf v^2 \: - u^2 = 2as \\ \\ :\implies \sf (0)^{2} - (20)^{2} = 2(-2)(s) \\ \\ :\implies \sf 0 - 400 = -4s \\ \\ :\implies \sf -400 = -4s \\ \\ :\implies \sf 400 = 4s \\ \\ :\implies \sf \dfrac{400}{4} \: = s \\ \\ :\implies \sf \cancel{\dfrac{400}{4}} \: = s \: \lgroup Cancelling \rgroup \\ \\ :\implies \sf 100 \: = s \\ \\ :\implies \sf s \: = 100 \: m \\ \\ :\implies \sf Distance \: = 100 \: metres

By using second equation of motion!

:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 20(10) + \dfrac{1}{2} \times (-2)(10)^{2} \\ \\ :\implies \sf s \: = 200 + \dfrac{1}{2} \times (-2)(100) \\ \\ :\implies \sf s \: = 200 + \dfrac{1}{2} \times -200 \\ \\ :\implies \sf s \: = 200 + 1 \times -100 \\ \\ :\implies \sf s \: = 200 + (-100) \\ \\ :\implies \sf s \: = 200 - 100 \\ \\ :\implies \sf s \: = 100 \: m \\ \\ :\implies \sf Distance \: = 100 \: m



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