Question

A bus is moving on a straight road with a speed of 126 kmh^-1 is brought to rest after covering a distance of 200m calculate retradation in cms^-2​

Answer 1

Answer:

as 126km/hr = 35m/sec{ initial speed}

as in the end bust will be in rest so final speed = 0

total distance = 200m

so use v²=u²+2as

so 0² =  35×35+ 2×200× a

-{35×35} ÷ 400 = a

so a = -3.06

as retradation is -ve{opposite} of accleration so

retardation = - {accleration}

                   = 3.06m/s²

as 1 meters per (second ²) =

100 centimeters per (second² )

so 3.06m/s² = 306cm/s²

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Answer 2

$u=\frac{126km}{h}$

$u=35ms^-1$

v=0

s=200m

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

$0=35^2+2×a×200$

$0=1225+400a$

$-1225=400a$

$a=\frac{-1225}{400}$

$a=-3.06$

$retardation\: is \:3.06ms^-2$