Question

A bus moving at a speed of 72km/h applies break and steadily retards at a rate of 2m/s.Find the time taken for it to halt.
Pls ans with complete steps
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Answer 1

Given :

▪ Initial speed of bus = 72kmph

▪ Retardation = -2m/s²

▪ Final speed of bus = zero

To Find :

▪ Time taken by bus to halt.

Concept :

✏ Since acceleration has said to be constant throughout the deceleratory period, we can easily apply equation of kinematics to solve this type of questions.

First equation of kinematics :

☞ v = u + at

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

Conversion :

◐ 1kmph = 5/18mps

◐ 72kmph = 72×5/18 = 20mps

Calculation :

→ v = u + (-a)t

[Negative sign shows retardation.]

→ 0 = 20 - 2t

→ 2t = 20

→ t = 10s

Answer 2

GIVEN

◙ Initial velocity of a bus(u) = 72km/h

◙ Retardation of the bus(a) = -2m/s²

◙ Final velocity of the bus(v) = 0 m/s

TO FIND

The time taken(t) for the bus to halt.

SOLUTION

Given, u = 72 km/h

Converting it into m/s :-

⇒u = 72 × 5/18

⇒u = 5 × 4

⇒u= 20 m/s

Using the formula :-

$\boxed{\color{purple} {\sf{v=u+at}}}$

Substituting the values :-

$\sf{\longrightarrow 0=20+(-2)t}\\\\\sf{\longrightarrow -20=(-2)t}\\\\\sf{\longrightarrow \dfrac{-20}{-2} =t}\\\\\large\underline{\boxed{\boxed{\color{green} {\sf{\longrightarrow 10\ s=t}}}}}$

Therefore, the time taken by the bus to halt is 10 seconds.