Question

Pl Solve the question ​

Answer 1

⇝ Given :-

A jeep travelling with a velocity of 108 km/h.

It is brought to rest by applying brakes.

It experiences a uniform retardation of 3 m/s²

⇝ To Find :-

Distance travelled by Jeep before coming to rest.

⇝ Solution :-

❒ Converting Initial Velocity in m/s ;

$\begin{gathered}\rm Initial \: Velocity = 108 \: km/h \\ \end{gathered} </p><p>$

$\begin{gathered} = \rm\bigg(108 \times \frac{5}{18} \bigg)m/s \\ \end{gathered} </p><p>$

$\begin{gathered} = \rm \bigg(\frac{540}{18} \bigg)m/s \\ \end{gathered}$

$\begin{gathered}:\longmapsto \red{ \rm Initial \: Velocity = 30 \: m/s } \\ \end{gathered}$

❒ Calculating Required Distance :

We Have,

Initial Velocity = u = 30 m/s

Final Velocity = v = 0 m/s

Acceleration = a = - 3m/s²

Let time taken by Jeep to stop be = t second.

★ We Have 1st Equation of Motion as :

$\begin{gathered}\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ v = u + at }}} \\ \end{gathered} </p><p>$

⏩ Putting Values In 1st Equation :

$\begin{gathered}:\longmapsto \rm 0 = 30 + ( - 3) \times t \\ \end{gathered}$

$\begin{gathered}:\longmapsto \rm \cancel - 3t = \cancel - 30 \\ \end{gathered}$

$\begin{gathered}:\longmapsto \rm t = \cancel \frac{30}{3} \\ \end{gathered}$

$</p><p>\purple{ \large :\longmapsto \underline {\boxed{{\bf t = 10 \: s} }}}$

★ We Have 2nd Equation of Motion as :

$\begin{gathered}\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}} \\\end{gathered} </p><p>$

⏩ Putting Values In 2nd Equation :

$\begin{gathered}:\longmapsto \rm s = 30 \times 10 + \frac{1}{2} \times ( - 3) \times {10}^{2} \\ \end{gathered} </p><p>$

$\begin{gathered}:\longmapsto \rm s = 300 - 150 \\ \end{gathered}$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf s = 150 \: m} }}}:⟼ </p><p> s=150m$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{Required \: Distance = 150 \: m}}}}} </p><p>$