The distance `x` covered by a particle varies with time `t` as `x^(2) = 2t^(2) + 6t + 1`. Its acceleration varies· with `x` as

Options

A. a) x
B. b)`x^2`
C. `c)
x^(-1)`
D. `d)x^(-3)`

Answer

D

Explanation

Given , `x^(2) = 2t^(2) + 6t +1" "...(i)`
Differentiating Eq. (i) w.r.t. t, we get
`2x (dx)/(dt) = 4t + t`
`2xv = 4t + 6 ( because " "v= (dx)/(dt))`
`xv = 2t + 3 " "...(ii)`
Now, again differentiating Eq. (ii) w.r.t. t, we get
`x (dv)/(dt) + v (dx)/(dt) =2`
`x. a + b. v=2 (because a= (dv)/(dt) and v= (dx)/(dt) )`
`xa + v^(2) =2 " "...(iii)`
Here, `v^(2) = (4t^2 + 12t + 9)/( x^2)`
`v^2 = (2 (2t^2 + 6t + 1) + 7)/( x^2)`
`v^2 = (2t^2 + 7)/( x^2) " "...(iv)`
Put the value of `v^2` in Eq. (iii) form Eq. (iv), we get
`xa + (2x^(2) +7)/( x^2) = 2`
`x^3 a + 7=0`
`x^(3) a =-7`
`a= (-7)/( x^3)`
Hence, the acceleration varies `x^(-3)`.