Question

a buss is moving with a velocity of 72kmh on the application of the breaks its stop after covering a distance of 500m calculate decelaration produced by the breaks

Answer 1

think the phrase “kmh raised to the power -1” means kilometers per hour. So the bus is decelerating from 72 kilometers/hour in the distance of 500 meters.

Deceleration is the same as acceleration in physics, and acceleration is (change in velocity)/(time). So the trick is to figure out the time. If it stopped in ten seconds, for example, the answer would be 72 km/hr/10 seconds. I would imagine that the teacher wants you to use cleaner units, so:

convert 72 km/hr into a velocity of units m/s. 1 km/hr = 1000 m/3600 seconds. For the sake of example, let’s pretend it is 10 m/s (that’s not the right answer, I am a typical teacher asking you to solve the math yourself).
assume that the acceleration is constant, so that =+
V=u t.+a t
where
v
o
= 10 m/s (again - that is not the real value). Since that is a “linear” equation, it means that the velocity goes linearly from 10 to 0 over time. That means we can use
v
o
/2 or in our example, 5 m/s, as the average velocity for the time of stopping.
calculate the time to go 500 meters using your average velocity. So in this case 500m/5m/s = 100 seconds. That’s the time we needed!
now acceleration = velocity change/time = 10 m/s / 100 seconds = 0.1 m/s2
That is a small acceleration. The initial velocity is greater than 10 m/s, so you will find that the acceleration is larger. Remember that gravity is 9.81 m/s2. Your result will be less than that 9.81 (which would be some great breaks!) but more than 0.1.

Answer 2

Given-

• Initial Velocity = 72km/h
• Final Velocity = 0
• Distance = 500 m

   

Find-

• Deceleration-?

   

Formula to be used-

• $\boxed{\sf{v^2-u^2= 2as}}$

where,

v= final velocity

u= initial velocity

a= acceleration or (-declaration)

s= distance

     

Solution-

• First change 72km/h to m/s

we know

1 km= 1000 m

and 1hr = 3600 s

so,

$\sf{1km/h = \dfrac{1000m}{3600s} = \dfrac{5m}{18s}}$

$\sf{\implies 72 km/h = \dfrac{5}{18} \times 72 m/s}$

Initial Velocity [u] = 20 m/s.

   

• Now By using 3rd eq' of motion,

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

put values,

$\bf{\implies 0^2 - 20^2 = 2\times a \times 500}$

$\bf{ \implies -400 = a\times 1000}$

$\bf{\implies a= \dfrac{-400}{1000} = \dfrac{-4}{10} }$

$\bf{\implies a= -0.4 }$

   

Therefore,  Deceleration of the bus is -0.4m/s².