Question

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4.0 m/s^2. Find the final velocity of the roller coaster ?​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ (u) = 13 \ m/s }$

$\:\:\:\:\bullet\:\:\:\sf{Distance \ (s) = 400 \ m }$

$\:\:\:\:\bullet\:\:\:\sf{ Acceleration = 4 \ m/s^2}$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{The \:Final \:velocity \:of \:the\: body }$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ Using 3rd equation of motion

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} +2as }$

$\dashrightarrow\:\: \sf{ {v}^{2} = {13}^{2} + 2 \times 4 \times 400 }$

$\dashrightarrow\:\: \sf{{v}^{2} = 169 + 3200 }$

$\dashrightarrow\:\: \sf{ {v}^{2} = 3369 }$

$\dashrightarrow\:\: \sf{ v = \sqrt{3369} }$

$\dashrightarrow\:\: \underline{\boxed{\sf{ v = 58.04 \: m/s }}}$

${\mathfrak{\underline{\purple{\:\:\:Additional \:Information:-\:\:\:}}}}$

☢ Equations Of Motion

$\boxed{

\begin{minipage}{3 cm} \\

\sf{\:\:\star\:\:v = u +at} \\ \\

\sf{\:\:\star\:\:s = ut + \dfrac{1}{2}\:at^{2} }\\ \\

\sf{\:\:\star\:\:v^{2} = u^{2} + 2as}\\ \\

\sf{\:\:\star\:\:s = \dfrac{1}{2} (u + v)t}\\

\end{minipage}

}$

$\sf{Where,}$

$\:\:\:\:\bullet\:\:\:\textsf{v = Final velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{u = Initial velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{a = Acceleration}$

$\:\:\:\:\bullet\:\:\:\textsf{s = Distance}$

$\:\:\:\:\bullet\:\:\:\textsf{t = Time taken}$