Question

2. A bus decreases its speed from 80 km h'1 to 60 km h-1 in 5 s. Find the acceleration of the
bus.​

Answer 1

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 80 km/h
• Final velocity (v) = 60 km/h
• Time taken (t) = 5s

We are asked to calculate the acceleration of the bus.

Acceleration is defined as rate of change in velocity with time. Its SI unit is m/s². So, we'll be calculating the acceleration in its SI unit.

Here, we can solve this question in two ways,

• By using the formula of acceleration.
• by using the first equation of motion.

But before commencing the steps, let's first convert the given velocities in m/s.

Converting initial velocity 80 km/h into m/s :

$\\ \longrightarrow \sf{\quad { 1 \; km/h = \dfrac{5}{18} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 80 \; km/h = 80 \times \dfrac{5}{18} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 80 \; km/h = 40 \times \dfrac{5}{9} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 80 \; km/h = \dfrac{200}{9} \; m/s }}$

$\\ \longrightarrow \bf{\quad \underline{80 \; km/h = 22.22 \; m/s }}$

Converting final velocity 60 km/h into m/s :

$\\ \longrightarrow \sf{\quad { 1 \; km/h = \dfrac{5}{18} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 60 \; km/h = 60 \times \dfrac{5}{18} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 60 \; km/h = 30 \times \dfrac{5}{9} \; m/s }}$

$\\ \longrightarrow \sf{\quad { 60 \; km/h = \dfrac{150}{9} \; m/s }}$

$\\ \longrightarrow \bf{\quad \underline{60 \; km/h = 16.66 \; m/s }}$

★ Calculating acceleration by using acceleration formula :

$\\ \longrightarrow \quad \pmb{\boxed{\sf {a = \dfrac{v -u}{t} }} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Substituting values,

$\\ \longrightarrow \sf{\quad {a = \dfrac{16.66 - 22.22}{5} \; m/s^2 }}$

$\\ \longrightarrow \sf{\quad {a = \cancel{\dfrac{-5.55}{5} }\; m/s^2 }}$

$\\ \longrightarrow \bf{\quad \underline{a= -1.11 \; m/s^2 }}$

★ Calculating acceleration by using the first equation of motion :

$\\ \longrightarrow \quad \pmb{\boxed{\sf { v = u + at}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Substituting values,

$\\ \longrightarrow \sf{\quad { 16.66 = 22.22 + 5a}}$

$\\ \longrightarrow \sf{\quad { 16.66 - 22.22= 5a}}$

$\\ \longrightarrow \sf{\quad { -5.55= 5a}}$

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

$\\ \longrightarrow \bf{\quad \underline{a= -1.11 \; m/s^2 }}$

Therefore, acceleration of the bus is -1.11 m/s².