Question

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

Answer 1

Going with the flow:

• Velocity :

`t=ax^2+bx`

`=> dt=2ax \ dx+b\ dx`

`:. qquad dx/dt=1/(2ax +b) qquad qquad qquad color(blue)[ = v equiv (dx)/(dt)]`

• Acceleration :

`(d^2 x)/(dt^2) = (dv)/dt = (dv)/(cancel(dx))(cancel(dx))/(dt) = d/(dx) (1/(2ax +b)) * 1/(2ax +b)`

`= - 2a (1/(2ax +b))^(2) * 1/(2ax +b)`

`:. (d^2 x)/(dt^2) = - (2a)/ ((2ax +b)^(3)) qquad qquad " which is acceleration"`

This solves the literal problem.