# The acceleration between time t and distance x is t=ax^2+bx where a and b areconstants.the acceleration is?

Aug 9, 2018

Going with the flow:

• Velocity :

$t = a {x}^{2} + b x$

$\implies \mathrm{dt} = 2 a x \setminus \mathrm{dx} + b \setminus \mathrm{dx}$

$\therefore q \quad \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2 a x + b} q \quad q \quad q \quad \textcolor{b l u e}{= v \equiv \frac{\mathrm{dx}}{\mathrm{dt}}}$

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} = \frac{\mathrm{dv}}{\mathrm{dt}} = \frac{\mathrm{dv}}{\cancel{\mathrm{dx}}} \frac{\cancel{\mathrm{dx}}}{\mathrm{dt}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{2 a x + b}\right) \cdot \frac{1}{2 a x + b}$

$= - 2 a {\left(\frac{1}{2 a x + b}\right)}^{2} \cdot \frac{1}{2 a x + b}$

$\therefore \frac{{d}^{2} x}{{\mathrm{dt}}^{2}} = - \frac{2 a}{{\left(2 a x + b\right)}^{3}} q \quad q \quad \text{ which is acceleration}$

This solves the literal problem.