Question

a bus is moving with a velocity of 30km/h. on applying the brakes it stops at a distance of 10m. calculate the time taken to stop the bus.​

Answer 1

Answer:

the bus comes to rest in 2/3 hrs

Explanation:

A bus initially moves with a velocity of 30km/hr right? So initial velocity or u = 30km/he.

Now , since the bus comes to rest or it stops after applying the brakes , so the final velocity or v of the bus is 0 km/hr .

We also know that after applying the brakes ,  

The bus travels a distance or S of 10 meters

So we can conclude that - " After applying the brakes , the bus undergoes retarding motion ie the speed of the bus decreases

And while the speed of the bus keeps on decreasing , the bus travels 10 m .  Therefore u = 300km/he

v = 0km/he

S = 10m

So now we can apply the equation of motion

v^2 = u^2 + 2aS .

Now we know the value of v , u and S .

So we can use the above equation to calculate acceleration or a .

After calculating the acceleration is found out to be -

v^2 = u^2 + 2aS

0^2 = 30^2 + 2×a×10

0 = 900 + 20a

-900 = 20a

-900/20 = a

a = -45km/he

Therefore acceleration is -45km/hr

Or retardation is 45km/hr

Since retardation = negative acceleration

now that we know v , u , S and a .

Let the time taken for the bus to stop be t hrs

Now

We can use the equation S = ut + 1/2 at^2

to calculate the time

the calculation will be -

10 = 30×t + 1/2 × -45 ×t^2

10= 30t - 45/2t^2

10 = ( 60t - 45t^2 ) / 2

20= 60t - 45t^2

45t^2 - 60t + 20 = 0

5( 9t^2 - 12t + 4 ) = 0

9t^2 - 12t + 4 = 0

9t^2 - ( 6 + 6 )t + 4 = 0

9t^2 - 6t - 6t + 4 = 0

3t( 3t - 2 ) - 2(3t - 2) = 0

(3t -2)(3t-2) =0

Therefore either 3t - 2 = 0

Or t = 2/3 hrs

Or 3t - 2 = 0

Or t = 2/3 hrs

Therefore the bus comes to rest in 2/3 hrs

Answer 2

Given:-

• Initial velocity = u = 30km/hour
• Final velocity = v = 0m/s
• Distance Travelled = 10m

━━━━━━━━━━━━━━━━

Need to find:-

• Time taken to stop = ?

━━━━━━━━━━━━━━━━

Solution:-

→ u = 30km/hour

→ u = 30 × 1000m/hour

→ u = 30000 ÷ 3600 m / sec

→ u = 50/6 m/sec

━━━━━━━━━

From 3rd equation of motion,

${v}^{2} - {u}^{2} = 2as$

Substitute the values,

→$0 - \dfrac{2500}{36} = 2 \times a \times 10$

→$20 \times a = - 69.44$

→$a = \dfrac{-69.44}{20}$

→$a = -3.47$

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We know from 1st equation of motion:

v = u + at

→ $0 = \dfrac{50}{6} - 3.47 \times t$

→$0 = 8.33 - 3.47 \times t$

→$- 3.47 \times t = - 8.33$

→$3.47 \times t = 8.33$

→$t = 8.33 \div 3.47$

→$\red{\boxed{\bold{\large{t \approx \: 2.4sec}}}}$