Question

A bus decreases its speed from 80 Km/h to 60 Km/h in 5s. Find the acceleration of the bus?

Answer 1

Answer:

u= initial velocity= 80 km/h = 22.2 m/s

v= final velocity = 60 km/h = 16.7 m/s

t= time = 5 s

a= acceleration = ?

a=v-u/t

a= 16.7-22.2/5

a= -5.5/5

Therefore, a=-1.1 m/s^2

Answer 2

Answer:

• The Acceleration (a) of the bus is - 1.12 m/s²

Given:

1. Initial velocity (u) = 80 Km/h
2. Final velocity (v) = 60 Km/h
3. time taken (t) = 5 sec.

Explanation:

$\rule{300}{1.5}$

Initial Velocity = ? m/s

As, the initial velocity is given in Km/h we need to convert it into m/s ,

Therefore,

From conversion factor we know,  

⇒ 80 Km / h = 80 × ( 5 / 18) m/s

⇒ 80 Km / h = 400 / 18

⇒ 80 Km / h = 22.2 m/s

⇒ Initial velocity (u) = 22.2 m/s.

Final Velocity = ? m/s

As, the Final velocity is given in Km/h we need to convert it into m/s ,

Therefore,

From conversion factor we know,  

⇒ 60 Km / h = 60 × ( 5 / 18) m/s

⇒ 60 Km / h = 300 / 18

⇒ 60 Km / h = 16.6 m/s

⇒ Final velocity (v) = 16.6 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the formula we know,

⇒ a = ( v - u ) / t

Where,

• a Denotes Acceleration.
• v Denotes Final velocity.
• u Denotes Initial velocity.
• t Denotes Time period.

Now,

⇒ a = ( v - u ) / t

Substituting the values,

⇒ a = ( 16.6 - 22.2 ) / 5

⇒ a = - 5.6 / 5

⇒ a = - 1.12

⇒ a = - 1.12 m/s²

∴ The Acceleration (a) of the bus is - 1.12 m/s².

$\rule{300}{1.5}$