Question

A bus moving with a velocity of 60 km h-1 is brought to rest in 20 seconds by applying brake find acceleration​

Answer 1

$\large \dag$ Question :-

A bus moving with a velocity of 60 km h-1 is brought to rest in 20 seconds by applying brake find acceleration 

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Acceleration   \:  is \:  4 \: m/s^2}} }$

$\large \dag$ Step by step Explanation :-

❒ Converting Initial Velocity in m/s :-

We Have,

$\text{Initial Velocity= 60 km/h}$

$= \rm\bigg( 60\times \frac{5}{18}\bigg)second$

$\red{:\longmapsto \text{Initial Velocity = 16.66 \:m/s}}$

Now Here we have :

• Initial Velocity = u = 16.66 m/s

• Final Velocity = v = 0 m/s

• Time = t = 20 s

• Let acceleration be = a m/s²

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{\blue{v = u + at}}}}$

⏩ Applying 1st Equation of Motion ;

$\\ :\longmapsto \rm 0 = 16.66 + a \times 20$

$:\longmapsto \rm - 16.66 = 20a$

$:\longmapsto \rm a = \frac{ - 16.66}{ \: \;\: 20}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = - 0.83} }}}$

⇝ Acclerration of bus is - 0.83 m/s²

Answer 2

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

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 ) .

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

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

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

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

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

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

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

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

• 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 .

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