Question

A bus decreases it's speed from 80 Km h^-1 to 60 km h^-1 in 5s. find the acceleration of the bus​

Answer 1

Answer:

Given :-

• A bus decreases it's speed from 80 km/h to 60 km/h in 5 seconds.

To Find :-

• What is the acceleration of the bus.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

Solution :-

First, we have to convert km/s into m/s :

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: initial\: velocity\: (u) \: :-}}}}}$

$:\implies \sf Initial\: Velocity =\: 80\: km/h$

$:\implies \sf Initial\: Velocity =\: 80 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$:\implies \sf Initial\: Velocity =\: \dfrac{400}{18}\: m/s$

$:\implies \sf\bold{\green{Initial\: Velocity =\: 22.22\: m/s}}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: final\: velocity\: (v)\: :-}}}}}$

$:\implies \sf Final\: Velocity =\: 60\: km/h$

$:\implies \sf Final\: Velocity =\: 60 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$:\implies \sf Final\: Velocity =\: \dfrac{300}{18}\: m/s$

$\implies \sf\bold{\green{Final\: Velocity =\: 16.67\: m/s}}$

Now, we have to find the acceleration :

Given :

• Final Velocity (v) = 16.67 m/s
• Initial Velocity (u) = 22.22 m/s
• Time (t) = 5 seconds

According to the question by using the formula we get,

$:\longrightarrow \sf 16.67 =\: 22.22 + a(5)$

$:\longrightarrow \sf 16.67 =\: 22.22 + 5a$

$:\longrightarrow \sf 16.67 - 22.22 =\: 5a$

$:\longrightarrow \sf \bigg(\dfrac{1667}{100}\bigg) - \bigg(\dfrac{2222}{100}\bigg) =\: 5a$

$:\longrightarrow \sf \bigg(\dfrac{1667 - 2222}{100}\bigg) =\: 5a$

$:\longrightarrow \sf \bigg(\dfrac{- 555}{100}\bigg) =\: 5a$

$:\longrightarrow \sf (- 5.55) =\: 5a$

$:\longrightarrow \sf - 5.55 =\: 5a$

$:\longrightarrow \sf \dfrac{- 5.55}{5} =\: a$

$:\longrightarrow \sf - 1.11 =\: a$

$:\longrightarrow \sf\bold{\red{a =\: - 1.11\: m/s^2}}$

$\therefore$ The acceleration of the bus is - 1.11 m/s².

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \dashrightarrow \: \underline{\large\boxed{ \red{ \rm \pmb Question :}}}$

• A bus decreases it's speed from 80 Km h^-1 to 60 km h^-1 in 5s. find the acceleration of the bus.

$\: \: \: \: \: \: \: \: \: \: \: \: \dashrightarrow \: \underline{\large\boxed{ \pink{ \rm \pmb Given :}}}$

• A bus decreases it's speed from 80 Km h^-1 to 60 km h^-1 in 5s

$\: \: \: \: \: \: \: \: \: \: \: \: \dashrightarrow \: \underline{\large\boxed{ \purple{ \rm \pmb To \: Find :}}}$

• find the acceleration of the bus

$\: \: \: \: \: \maltese \: \large{\underline{\underline{ \sf \blue {Using \: formula : }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \dag \: {\boxed{\boxed{\red{ \bf \: v = u + at}}}} \: \dag$

Where as :

$\: \: \: \: \: :\rightarrow {\underline{\underline { \rm \: v = final \: velocity}}}$

$\: \: \: \: \: :\rightarrow {\underline{\underline{{ \rm \: u = initial \: velocity}}}}$

$\: \: \: \: \: :\rightarrow {\underline{\underline{ \rm \: a = acceleration}}}$

$\: \: \: \: \: :\rightarrow {\underline{\underline{ \rm \: t = time}}}$

Convert km/s into m/s :

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm \: u = 80 km/hrs$

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm u = \: 80 \times \dfrac{5}{18} \: m/s$

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm u = \dfrac{400}{18} m/s$

So in decimals the u is = 22.22 m/s.

$\: \: \: \: \: \: \: \: \: \: \: \: \maltese \: \: {\underline{\underline{ \rm \: So \: now \: find \: v}}}$

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm v =60km/hrs$

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm v = 60× \dfrac{5}{18} m/s$

$\: \: \: \: \: \: \: \: \: \: : \: \longmapsto \rm v = \dfrac{300}{18} m/s$

So in decimals the v is = 16.67 m/s.

Calculation :

• v = 16.67 m/s

• u = 22.22 m/s

• t = 5 s

$\: \: \: \: \: \: \: \: \: \: : \rightarrow \rm \: v = u + at$

$\: \: \: \: \: \: \: \: \: \: : \rightarrow \rm 16.67 = 22.22 + a(5)$

$\: \: \: \: \: \: \: \: \: \: : \rightarrow \rm 16.67= 22.22+5a$

$\: \: \: \: \: \: \: \: \: \: : \rightarrow \rm 16.67 - 22.22 = 5a$

$\: \: \: \: \: \: \: : \rightarrow \rm \bigg \{\dfrac{1667}{100} - \dfrac{2222}{100} \bigg \}= 5a$

$\: \: \: \: \: \: \: \: : \rightarrow \rm \bigg \{\dfrac{1667 - 2222}{100} \bigg \}= 5a$

$\: \: \: \: \: \: \: \: \: \: : \rightarrow \rm \bigg \{\dfrac{ - 555}{100} \bigg \}= 5a$

$\: \: \: \: \: \: \: \: \: \: \: : \rightarrow \rm \big\{ - 5.55\big\}= 5a$

$\: \: \: \: \: \: \: \: \: \: \: \: \: : \rightarrow \rm a = \dfrac{ - 5.55}{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: : \rightarrow \rm a = - 1.11 \: m/s^2$

$\therefore \rm {\underline{\underline{The \: acceleration \: of \: bus \: is \: 1.11 \: m/s^2}}}$