The relation between time t and distance x is t=αx2+βx . . .

Question :  The relation between time t and distance x is t=αx²+βx where α and β are constants. Find velocity and acceleration.

Doubt by Zoha

Solution : 

t=αx²+βx

differentiating both sides w.r.t. x

d(t)/dt = d(αx²+βx)/dt
1=d(αx²)/dt+d(βx)/dt

1=2αx(dx/dt)+β(dx/dt)

1=2αxv+βv [∵dx/dt=v]

1=(2αx+β)v
v=1/2αx+β — (1)

v=(2αx+β)^-1
Again differentiating w.r.t. time both sides

dv/dt = (-1)(2αx+β)^-2×d(2αx+β)/dt
a = -(2αx+β)^-2×[2α(dx/dt)+0]
a = -(2αx+β)^-2×[2αv]
a = -2αv/(2αx+β)²
a = -2αv×[1/2αx+β]²
a = -2αv×[v]² [From equation (1)]
a = -2αv³

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Similar Question : The relation between time t and distance x is t=ax²+bx where a and b are constants. The acceleration is [AIEEE 2005]

a) 2bv³

b) -2abv²

c) 2av²

d) -2av³