Question

A bus is running with a velocity of 60km/hr.On seeing a child at 11m ahead on the road,the driver applied brakes and the bus stops at distance of 10m.Calculate retardation .

Answer 1

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✤ Required Answer:

✒ GiveN:

• Initially velocity was 60 km/hr
• Distance covered before stopping = 10m

✒ To FinD:

• Retardation of the car.....?

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✤ How to solve?

In order to solve this question, we need to know the three equations of motion which is based on speed, velocity, distance, time and acceleration.

Three equations of motion,

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

⚠️ These equations are used when the body is under uniform acceleration or no acceleration. For Non-uniform acceleration, we use Calculus..

So, Let's solve this question, by using these three equations of motion$.$

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✤ Solution:

We have,

• Initial velocity(u) = 60 km/hr

[ Unit conversion km/hr ➝ m/s ]

➝ Initial velocity in m/s = 60 × 5/18

➝ Initial velocity in m/s = 50/3 m/s

• Distance covered(s) = 10 m
• Final velocity(v) = 0 m/s

[Finally the car stops, So v = 0]

So, By using 3rd equation of motion,

➙ v² = u² + 2as

➙ 0² = (50/3)² + 2 × a × 10

➙ 0 = 2500/9 + 20a

➙ 20a = -2500/9

➙ a = -125/9 = -13.89 m/s

☕ We know, Retardation = Negative Acceleration

➙ Retardation = -(-13.89) m/s

➙ Retardation = 13.89 m/s

✒ Therefore, Retardation of the car = 13.89 m/s

Hence, Solved !!

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Answer 2

Given :

• Initial Velocity [u] = 60 km/h

We will convert km/h to m/s...

➺ 60 × $\dfrac{5}{18}$

➺ $\dfrac{50}{3}$

• Distance covered[s] = 10 m
• Final Velocity [v] = 0 m/s

To find :

• Retardation

Formula used :

➺ v² = u² + 2as

➺ 0² = $\sf(\dfrac{50}{3})^{2}$+ 2 × a × 10

➺ 0 = $\sf\dfrac{2500}{9}$ + 20 a

➺ 20 a = $\sf\dfrac{- 2500}{9}$

➺ a = $\sf\dfrac{-125}{9}$

➺ a = - 13.89 m/s.

.°. Retardation = 13.89 m/s...