The relation between time ‘t’ and distance ‘x’ is αx^2 + βx, where α and β

Question

The relation between time ‘t’ and distance ‘x’ is \(\alpha x ^2 +\beta x, \) where \(\alpha\) and \(\beta\) are constants. The relation between acceleration (a) and velocity (v) is

(1) \(a = -2\alpha v^3\)

(2) \(a = -5\alpha v^5\)

(3) \(a = -3\alpha v^2\)

(4) \(a= -4\alpha v^4\)

Answer 1

(1) \(a = -2\alpha v^3\)

\(t = \alpha x ^2 +\beta x\) (differentiating wrt time)

\(\frac {dt}{dx}= 2 \alpha x + B\)

\(\frac {1}{v} = 2\alpha x +\beta\)

(differentiating wrt time)

\(-\frac {1}{v^2}\frac {dv}{dt}= 2\alpha \frac {dx}{dt}\)

\(\frac {dv}{dt}=-2\alpha v^3\)