Question

A bus is moving on a straight road with a speed of 126 km h-1 is brought to rest after 200 m. Calculate retardation of the bus.​

Answer 1

Answer:

3.062 $m/sec^2$

Explanation:

Given :

Initial velocity ( u ) = 126 km / hr

To convert km / hr into m / sec multiply it by 5 / 18 we get

u = 7 × 5 = 35 m / sec

Final velocity ( v ) = 0 m / sec

Distance ( s ) = 200 m

From third equation of motion we have

$\displaystyle v^2=u^2+2as$

where v  = final velocity  

u = Initial velocity    

a = Acceleration

s = Distance

Put the values in equation we get

$\displaystyle 0^2=(35)^2+2\times200\times a\\\\\displaystyle 0=1,225+400\times a\\\\\displaystyle 400\times a=-1,225\\\\\displaystyle a=-\frac{1225}{400}\\\\\displaystyle a=-3.062 \ m/sec^2$

Negative acceleration means  acceleration in opposite direction also called retardation.

Thus we get retardation is 3.062 $m/sec^2$

Answer 2

» A bus is moving on a straight road with a speed of 126 km/h.

Here..

Initial velocity (u) = 126 km/hr

=> 126 × $\dfrac{5}{18}$

=> 35 m/s

» The bus is brought to rest after 200 m.

Here..

Final velocity (v) = 0m/s

Distance (s) = 200m

_________ [ GIVEN ]

• We have to calculate it's retardation (-a)

_____________________________

We know that ..

v² - u² = 2as

(Third Equation of Motion)

=> (0)² - (35)² = 2(a)(200)

=> - 1225 = 400a

=> 400a = - 1225

=> a = - 3.0625 m/s²

_____________________________

The retardation is 3.0625 m/s²

_________ [ ANSWER ]

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