a bus moving with the velocity 30km/h. on applying breaks it comes to rest after travelling a distance 10m.find time taken by the bus in coming to rest
Answers
Answer: Hey Buddy , here's the answer you're looking for -
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
Hope it helps you
Explanation:
Answer:
Explanation:
The unit of distance and initial velocity is different so we need to change it into same unit . The answer is in attachment and is 2.4 sec and in hr it will be 0.000667 hr .
