Physics, asked by yeasmin2814, 9 months ago

A bus moving with a velocity of 60km /h is brought to rest in 20 seconds by applying brakes .find the acceleration.

Answers

Answered by ashwinh2002
3

Answer:

Acceleration = - 0.85 m/s^{2} (Deceleration)

Explanation:

u = 60 Km/h = 60 x 5/18 = 16.66 m/s (approx , 17 m/s)

t = 20 s

v = 0 m/s (bus comes to rest)

First law of motion : v = u + at

a = v - u / t

  =  - 17 /20 =  - 0.85 m/s^{2}

Answered by SparklingThunder
0

\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}

A bus moving with a velocity of 60 km/h is brought to rest in 20 seconds by applying brake find acceleration ?

\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}

\textsf{Acceleration = $ \sf - 0.83 \: m {s}^{ - 2}  $}

\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}

\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}

  • Initial Velocity ( u ) = 60 Km/h

  • Final Velocity ( v ) = 0 m/s ( At rest )

  • Time taken ( t ) = 20 s

\green{ \large \underline{ \mathbb{\underline{TO  \: FIND : }}}}

  • Acceleration of bus ( a ) .

\green{ \large \underline{ \mathbb{\underline{ EQUATION \:  OF  \: MOTION \: USED : }}}}

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

\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}

  \red{\textsf{ \underline{\underline{Converting velocity into m${ \sf s}^{ - 1} $  : }}}}

  \begin{array}{l} \hline  \\ \small \displaystyle \sf \longrightarrow u =  \frac{60  \:km}{ h}  \times   \frac{1000 \: m}{1 \:  km}  \times  \frac{1  \: h}{60 \:min}  \times  \frac{1 \:min}{60  \:s}  \\  \\\small \displaystyle \sf \longrightarrow u =  \frac{600 \cancel{00 }\:m}{36 \cancel{00} \:s} \\  \\  \small \displaystyle \sf \longrightarrow u =  \frac{600 \:m}{36  \: s}   \\  \\\small \displaystyle \sf \longrightarrow u = 16.67 \: m {s}^{ - 1}    \\ \\  \hline \end{array}

 \begin{array}{l}  \hline \\  \displaystyle \sf \longrightarrow v = u + at  \\  \\ \displaystyle \sf \longrightarrow 0 =16.67 + a(20)   \\  \\ \displaystyle \sf \longrightarrow 0 =16.67+ 20a \\  \\ \displaystyle \sf \longrightarrow 16.67 + 20a = 0 \\  \\ \displaystyle \sf \longrightarrow 20a =  - 16.67  \\  \\ \displaystyle \sf \longrightarrow a =  \frac{ - 16.67}{20}   \\  \\ \displaystyle \sf \longrightarrow a =  - 0.83 \: m {s}^{ - 2}  \:  \:  \:  \:  \:  \:  \: \:\:\: \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\ \\    \hline  \end{array}

 \purple{ \boxed{ \textsf{Acceleration of bus} \sf  \:= -0.83  \: m {s}^{ - 2}   \:  \:  \:  \: }}

\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}

  • Acceleration

Acceleration is the rate at which velocity changes with time . Acceleration can be negative , positive or zero . Negative acceleration is called retardation .

  • Initial Velocity

Initial velocity is the velocity of the object before the effect of acceleration .

  • Final Velocity

Final velocity is the velocity of the object after the effect of acceleration .

   \Large{\purple{\boxed{\begin{array}{l} \textsf{Equations of motion : } \\  \\  \textsf{v = u + at} \\  \\   \displaystyle\textsf{s = ut +  $ \sf\frac{1}{2}a {t}^{2} $ } \\  \\ \sf  {v}^{2} -  {u}^{2}  =  2as \end{array}}}}

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