Question

A bus is moving with the velocity 30 kmh on applying the break it's comes to rest after traveling a distance 10 m. calculate the time taken by the bus in coming to rest​

Answer 1

Answer:

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 -

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

$0^{2} = 30^{2} + 2$×$a$×$10$

$0 = 900 + 20a-900 = 20a$

$\frac{-900}{20}= a$

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

To calculate the time

The calculation will be -

10 = 30×t+$\frac{1}{2}$×$-45$×$t^2$

$10= \frac{30t - 45}{2t^2}$

$10 = \frac{( 60t - 45t^2 )}{2}$

$20 = 60t - 45t^2$

$45t^2 - 60t + 20 = 0$

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

$9t^2 - 12t + 4 = 0$

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

$9t^2 - 6t - 6t + 4 = 03t( 3t - 2 ) - 2(3t - 2) = 0$

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

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