Question

A bus decreases it's speed from 80 km/h to 60 km/h in 10 seconds find the acceleration of the bus​

Answer 1

Given :

• Initial Velocity (u) = 80 km/h
• Final Velocity (v) = 60 km/h
• Time Taken (t) = 10 sec .

To Find :

• Acceleration of the bus .

Solution :

$\longmapsto\tt{Initial\:Velocity(u)=80\times\dfrac{5}{18}=22.22\:m/s}$

$\longmapsto\tt{Final\:Velocity(v)=60\times\dfrac{5}{18}=16.67\:m/s}$

Using 1st Equation :

$\longmapsto\tt\boxed{v=u+at}$

Putting Values :

$\longmapsto\tt{16.67=22.22+a(10)}$

$\longmapsto\tt{16.67-22.22=10a}$

$\longmapsto\tt{-5.55=10a}$

$\longmapsto\tt{a=\cancel\dfrac{-5.55}{10}}$

$\longmapsto\tt\bf{a=-0.55\:{m/s}^{2}}$

So , The Acceleration of the bus is -0.55 m/s² .

_______________________

Three Equations Of Motion :

• v = u + at
• s = ut + 1/2 at²
• v² - u² = 2as

Here :

• v = Final Velocity
• u = Initial Velocity
• t = Time Taken
• a = Acceleration
• s = Distance Travelled

_______________________

Answer 2

• $\boxed{\sf{\green{Acceleration\:(a) = \bf{-0.556\:m/s^{2}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Explanation :

$\begin{gathered}\underline{\bf\red{\bigstar}{\underline{Given}}}\begin{cases} & \sf{Initial\:velocity\:(u)\:of\:a\:bus = \bf{80\;km/h}} \\ \\ & \sf{Final\:velocity\:(v)\:of\:a\:bus = \bf{60\:km/h}}\\ \\ & \sf{Time\:taken\:(t)\:= \bf{6\;s}} \end{cases} \end{gathered}$

$\underline{\bf\red{\bigstar}{\underline{To\:Find}}:-}$

• Acceleration (a) of a bus = ?

$\underline{\bf\red{\bigstar}{\underline{Solution}}:-}$

• Here, we have initial velocity (u), final velocity (v) and time taken (t) by a bus. Firstly, we will convert units of initial velocity (u) and final velocity (v) from km/h to m/s. Then, by putting all values in first equation of motion, we will get required acceleration.

★ Converting units of initial velocity into m/s :-

$\qquad\leadsto\quad\sf Initial\:velocity = \bigg(80\:\times\:\dfrac{5}{18}\bigg)\:m/s$

$\qquad\leadsto\quad{\bf{\pink{ Initial\:velocity = 22.22\:m/s}}}$

★ Converting units of Final velocity into m/s :-

$\qquad\leadsto\quad\sf Final\:velocity = \bigg(60\:\times\:\dfrac{5}{18}\bigg)\:m/s$

$\qquad\leadsto\quad{\bf{\red{ Final\:velocity = 16.66\:m/s}}}$

★ Using first eqⁿ of motion to find "a" :-

$\qquad\longrightarrow\quad\sf v = u + at$

$\qquad\longrightarrow\quad\sf 16.66 = 22.22 + a(10)$

$\qquad\longrightarrow\quad\sf 16.66 - 22.22 = 10a$

$\qquad\longrightarrow\quad\sf -5.56 = 10a$

$\qquad\longrightarrow\quad\sf \dfrac{-5.56}{10} = a$

$\qquad\longrightarrow\quad\sf \dfrac{\cancel{-5.56}}{\cancel{10}} = a$

$\qquad\longrightarrow\quad{\bf{\purple{a = -0.556\:m/s^{2}}}}$

$\therefore\:{\underline{\sf{Hence,\:acceleration\:(a) = \bf{-0.556\:m/s^{2}}}}}$

$\underline{\bf\red{\bigstar}{\underline{More\:to\:know}}:-}$

★ Three equations of motion :-

• v = u + atㅤㅤㅤㅤㅤ[Used above]
• s = ut + (1/2) at²
• v² = u² + 2as

━━━━━━━━━━━━━━━━━━━━━━━━━━