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A car was movig at a rate of 18 kmh^(-1)...

A car was movig at a rate of `18 kmh^(-1)` . When the brakes were applied, it comes to rest in a distance of 100 m. Calculate the retardation produced by the brakes.

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Given, `v=O, u=18 kmh^(-1)=5ms^(-1),s=100 m`
Using the equation of motion
`v^(2)-u^(2)=2as` …..(i)
`therefore -u=2as " " (because v=0)`
`rArr s=-(u^(2))/(2a)rArr a=(-5xx5)/(2xx100)=-(1)/(8)=-0.125 ms^(-2)`
So, the retardation produced by the breakes is `0.125 ms^(-2)` .
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