The relation between time t and distance x is t = ax^(2)+ bx where a and b are constants. The acceleration is
Solution:-
Given that t = ax^(2)+ bx (where a and b are constant)
We know that for finding the acceleration, distance is differentiated double w.r.t time
dt/dt =d(ax^(2)+ bx)/dt
1=d(ax^2)/dt +dbx/dt
1=2ax.dx/dt+b.dx/dt
1=2ax.v+b.v
1=v(2ax+b)
2ax+b=1/v
Again diff, wrt time
d(2ax+b)/dt=d(1/v) /dt
d2ax/dt+db/dt=-(1/v^2).dv/dt
2a(dx/dt) +0=-(1/v^2) . A
2av=-1/v^2
-2av^3=A
A=-2av^3
Where A is accelaration.
