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A taxi without any passengers, moving on...

A taxi without any passengers, moving on a frictionless horizontal road, with a velocity u can be stopped in a distance d. Now the passengers and 40% to its weight. What is the stopping distance at veolcity u, if the retardation remains the same?

A

`sqrt(1.4d)`

B

`(1.4)^(2)d`

C

`1.4d`

D

`(1/1.4)d`

Text Solution

Verified by Experts

The correct Answer is:
C
Given : `m_(1)= m and m_(2)=1.4m`
`because v^(2)=u^(2)+2as " " therefore 0=u^(2)-2ad" " therefore a= (u^(2))/(2d)`
`therefore ` Retarding force `F= ma= (m u^(2))/(2d)`
`therefore d= (mu^(2))/(2F)" " ...(1)`
and in the second case, `m_(2)= 1.4 m`
` and d'=(1.4m u^(2))/(2F)" "...(2)`
`therefore (d')/(d)= 1.4 or d'= 1.4d`
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