Question

a bus decreases its speed from 80 kilometre per hour to 60 km per hour in 5 second find the acceleration of the bus.
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Answer 1

Answer:

-1.11 m/s²

Explanation:

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

• Initial velocity of the bus (u) = 80 km/h
• Final velocity of the bus (v) = 60 km/h
• Time taken (t) = 5 seconds

We are asked to calculate the acceleration of the bus.

Here, we can solve this question in two ways, by using the formula of acceleration and by using the first equation of motion. Before commencing the steps, we need to convert velocities in its Standard form.

★ Converting initial velocity in m/s :

• In order to convert km/h to m/s, we multiply the value with 5/18.

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

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

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

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

★ Converting final velocity in m/s :

• In order to convert km/h to m/s, we multiply the value with 5/18.

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

★ Converting initial velocity in m/s :

In order to convert km/h to m/s, we multiply the value with 5/18.

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

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

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

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

Calculating acceleration :

By using the 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

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

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

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

Therefore, acceleration is -1.11 m/s².

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

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

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

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

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

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

Therefore, acceleration is -1.11 m/s².