The relation between time ‘t’ and distance ‘x’ is t=ax^2+bx

The relation between time ‘t’ and distance ‘x’ is t=ax^2+bx,   where ‘a’ and  ‘b’ are constant. The acceleration is…..

a.   -2abv^2         
b.     2bv^2           
c.   -2av^3             
d.    2av^2

Solution:

We have,

t=ax^2+bx

Differentiating with respect to ‘t’

(dt/dt)=(d/dt)(ax^2+bx)

1=(d/dt) (ax^2 )+(d/dt)(bx)

1=a(d/dt) (x^2 )+b(d/dt)(x)

1=2 a x(  dx/dt)+b (dx/dt)

1=2 a x v+b v

1=2 a x v+b v

1=(2 a x +b)v

Therefore,

v=1/((2 a x +b) ) 

- - - - - - - - - -  (1)

W. k. t.

acceleration=dv/dt = d/dt (1/((2 a x +b) )  )

acceleration= ((-2a)/(2 a x +b)^2 ) dx/dt

acceleration= ((-2a)/(2 a x +b)^2 )  dx/dt

  acceleration=-2a(1/(2 a x +b)^2 )  v

 acceleration=-2a (〖 (1/((2 a x +b) )   )〗^2  )v

From equation (1)

acceleration=-2a(〖 v〗^2  ). v

Therefore,

acceleration=-2av^3

Answer :                                Option (c) is correct