1.

The relation between time `t` and distance `x` is `t = ax^(2)+ bx` where `a and `b` are constants. The acceleration isA. ` 2 bv^(3)`B. ` - 2 abv^(2)`C. ` 2 av^(2)`D. ` -2 av^(3)`

Answer» Correct Answer - D
`t = ax^(2) + bx` ,Diff. with repect to time `(t)`
` (d)/(dt)(t) = a(d)/(dt) (x^(2)) + b(dx)/(dt) = a.2x (dx)/(dt) + b.(dx)/(dt)`
` 1 = 2axv + bv = v( 2ax +b) rArr 2ax +b = 1/v`.
Again differentating , ` 2a(dx)/(dt) + 0 = -1/v^(2) (dv)/dt) `
`rArr (dv)/(dt) = f = -2av^(3)` `(because (dv)/(dt) = f = acc)`


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