Math, asked by gloriaekka948, 1 month ago

racing car at rest accelerates uniformly to a velocity of 144 km per hour in 20 seconds find the value of acceleration of the car​

Answers

Answered by chirrasrinivas
0
  • Step-by-step explanation:

40m/sec/20 sec =2m/sec2(square)

Attachments:
Answered by Anonymous
9

\boxed{ \underline{ \underline{\bigstar\tt\purple{  \: Solution:-}}}}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt \purple{Given:-}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{u = 0 \: (Because \:  car \:  is  \: at  \: rest)}

\footnotesize\tt{v = 144 \: km \: per \: hour}

\footnotesize\tt{t=20 \: sec}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt \purple{Now, convert  \: v  \: in \:  \:  m  \: per  \: sec}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{v =  \cancel{144}  \:  \times \: } \tt{ \frac{5}{ \cancel{18}} \:  \: m \: per \: sec }

\footnotesize\tt{v =  8  \:  \times 5 \:  m \: per \: sec }

\footnotesize\tt{v = 40\:  m \: per \: sec }

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt \purple{Now, find \: Acceleration \: (a)}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{v = u + at}

\footnotesize\tt{40= 0 + a \times 20}

\footnotesize\tt{40= 0 + 20a}

\footnotesize\tt{40 - 0=  20a}

\footnotesize\tt{40 =  20a}

\tt{ \frac{40}{20}}  \:  \footnotesize \tt{=  a}

\tt{  \cancel\frac{40}{20}}  \:  \footnotesize \tt{=  a}

\boxed{ \underline{ \underline{ \footnotesize \tt \purple{a = 2 \: m {s}^{ </strong><strong>2</strong><strong>}</strong><strong> }}}}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\</strong><strong>f</strong><strong>o</strong><strong>o</strong><strong>t</strong><strong>n</strong><strong>o</strong><strong>t</strong><strong>e</strong><strong>s</strong><strong>i</strong><strong>z</strong><strong>e</strong><strong>\</strong><strong>t</strong><strong>t</strong><strong>{</strong><strong>Extra:</strong><strong>-</strong><strong>}</strong><strong>

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt \purple{Now, find \:Distance \: (s)}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{s= ut + \frac{1}{2}  a  {t}^{2}  }

\footnotesize\tt{s= 0 \times 20+ \frac{1}{2}  \times 2 \times {20}^{2}  }

\footnotesize\tt{s=0 \: +  \: } \tt{\frac{1}{ \cancel2}  \: } \footnotesize\tt{  \times  \: \cancel 2 \times {20}^{2}  }

\footnotesize\tt{s=0  \:  +  \:  {20}^{2}   }

\footnotesize\tt{s=0  \:  +  \:  400 }

\boxed{ \underline{ \underline{\footnotesize\tt \purple{s= 400 \:  m}}}}

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