Question

A bus moving with a velocity of 60km/hr is brought to rest in 20 second by applying brakes. Find its acceleration

Answer 1

Answer :

Given,

Initial velocity, U = 60km/hr

$= > 60 \times \frac{5}{18} \\ \ = > 16.7m {s}^{ - 1}$

Final velocity,V= 0

Time = 20s

Acceleration, a =

$a = > \frac{v - u}{t} \\ = > \frac{0 - 16.7}{20} \\ \\ = > \frac{16.7}{20} \\ \\ = > 0.835m {s}^{ - 2}$

Thanks

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 ) .

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

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

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

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

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

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

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

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

• 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 .

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