Physics, asked by kr942077, 7 months ago

A bus is moving with a speed of 54 km/have. on applying breaks, it stooped in 8 seconds. calculate the acceleration and distance travelled in this time​

Answers

Answered by Cosmique
14

GiveN :-

  • initial velocity of bus, u = 54 km h⁻¹
  • time taken to stop , t = 8 sec
  • final velocity , v = 0  

To FinD :-

  • Acceleration , a = ?
  • distance traveled in this time ,s =?

FormulaE :-

First equation of motion

\boxed{\sf{v=u+at}}

( where v is the final velocity , u is the initial velocity , a is the acceleration and t is the time taken )

Second equation of motion

\boxed{\sf{s=ut+\frac{1}{2}at^2}}

( where s is the distance traveled , t is the time , u is the initial velocity , a is the acceleration )

SolutioN :-

Converting initial velocity given in km h⁻¹ into m s⁻¹

initial velocity , u = 54 km h⁻¹

\sf{\purple{initial\;velocity\:,u=54\times\frac{5}{18}=15\;ms^{-1}}}

Finding acceleration

Using first equation of motion

\longmapsto\sf{v=u+at}

putting known values

\longmapsto\sf{0=15+a(8)}\\\\\longmapsto\underline{\boxed{\large{\sf{\purple{a=\frac{-15}{8}\;ms^{-2}}}}}}

Finding distance traveled after applying break

Using second equation of motion

\longmapsto\sf{s=ut+\frac{1}{2}at^2}

putting known values

\longmapsto\sf{s=15(8)+\frac{1}{2}(\frac{-15}{8})(8)^2}

\\\\\longmapsto\underline{\boxed{\large{\sf{\purple{s=60\;m}}}}}


RvChaudharY50: Perfect. ❤️
Answered by MystícPhoeníx
194

_______________________________

Question :-

▪ A bus is moving with a speed of 54 km/have. on applying breaks, it stooped in 8 seconds. calculate the acceleration and distance travelled in this time.

_______________________________

Given:-

  • Initial velocity (u) =54km/h=54×5/18=15m/s

  • Final velocity (v) =0m/s

  • Time taken =8s

To Find :-

  • Acceleration

  • Distance travelled

Solution :-

By using 1st equation of motion

➦ v =u+at

➭ 0=15+a ×8

➭ a = -15/8

➭ a = -1.875

here, negative symbol show retardation.

∴The acceleration of bus 1.875m/s².

And,

By using 2nd equation of motion

➦ s =ut +1/2at²

➭ s = 15×8+1/2 ×1.875× 8× 8

➭ s= 7200m

∴The distance covered by bus 7.2km


RvChaudharY50: Awesome .❤️
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