1.

A train is travelling aur a speed of 90km/h. Brakes are applied so as to produced a uniform acceleration of -0.5m/s. Find how far the train will go before it is brought to rest.Tomorrow is my exam and I want the solution

Answer»

\textbf{\underline{\underline{According\:to\:the\:Question}}}

★Given :-

a = -0.5m/s²

u = 90 km/hr

★Convert it into m/s

\tt{\rightarrow\dfrac{90\times 1000m}{60\times 60}=25m/s}

v = 0 m/s

★To Find :-

s = ?

★Solution :-

\textbf{\boxed{Third\; Equation\;of\;Motion}}

{\boxed{v^{2} = u^{2} + 2as}}

★We can ALSO write :-

{\boxed{s=\dfrac{v^{2}-u^{2}}{2a}}}

★Substitute the VALUES we get :-

\tt{\rightarrow a =\dfrac{(0)^{2}-{25}^{2}}{2\times(-0.5)}}

\tt{\rightarrow a =\dfrac{-25\times {25}^2}{-2\times 0.5}}

\tt{\rightarrow a =\dfrac{625}{1}}

= 625 m

\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ $\ Distance = Speed\times Time \\ \\ Displacement=Velocity\times time \\ \\ Average\; Speed = \dfrac{Initial\:\:Speed+Final\:\:Speed}{2} $\end{minipage}}


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