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

1 Answer
Aug 9, 2018

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)]#

#(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.