Question

a bike riding at 22.4m/s skids to come halt in 2.55sec. conclude the sikding distanance of bike

Answer 1

hi,

A biker is traveling at a speed of 22.4 m/s and skids to a stop in 2.55 s. So we have following information:

Initial speed of the bike i.e u= 22.4 m/s

Final speed of the bike i.e. v=0 m/s ( because car stops)

Time taken to come to stopping position t= 2.55 sec

We wish to determine an acceleration (which is in fact a retardation) a=?

We have equation of the motion

`v=u+at`

`0=22.4+2.55xxa`

`-2.55a=22.4`

`a=22.4/(-2.55)`

`=-8.784 ms^(-2)`

`v^2=u^2+2as`

`0=(22.4)^2+2xxsxx(-8.784)`

`-501.76=-17.568s`

`s=28.56 m`

Before bike stops it will travel 28.56 m.

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