A train moving with a velocity of $48 k m h^{- 1}$ is brought to rest in 11 second to avoid collision. What is the retardation of the train in $m s^{- 2}$ and $k m h^{- 2}$

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Solution

Retardation = speed / time

Retardation in $k m / h^{- 2}$

$= \frac{48}{\frac{11}{60 \times 60}}$

$= \frac{48}{0.003} = 16000 k m / h^{- 2}$

retardation in $m / s^{- 2}$

$= \frac{48 \times \frac{1000}{3600}}{11}$

$= \frac{13.33}{11} = 1.2 m / s^{- 2}$

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A train moving with a velocity of 48 km h−1 is brought to rest in 11 second to avoid collision. What is the retardation of the train in ms−2 and km h−2

Retardation = speed / timeRetardation in km/h−2=481160×60=480.003=16000km/h−2retardation in m/s−2=48×1000360011=13.3311=1.2m/s−2

A train moving with a velocity of $48 k m h^{- 1}$ is brought to rest in 11 second to avoid collision. What is the retardation of the train in $m s^{- 2}$ and $k m h^{- 2}$

Retardation = speed / time

Retardation in $k m / h^{- 2}$

$= \frac{48}{\frac{11}{60 \times 60}}$

$= \frac{48}{0.003} = 16000 k m / h^{- 2}$

retardation in $m / s^{- 2}$

$= \frac{48 \times \frac{1000}{3600}}{11}$

$= \frac{13.33}{11} = 1.2 m / s^{- 2}$