Question

body moving in a trip at 24 km/ h then n what will be its aceeleration if break are applied for 5.7895 mili second ​

Answer 1

Answer:

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 2/3p

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