"Zero Displacement at Time t=? - Solve for t

[DrunkEngineer]

Homework Statement

A body experiences acceleration "a" given by the expression a=At-Bt^2 where A and B are constants and t is time. If at time t=0, the body has zero displacement and velocity, at what next value of time does the body again have zero displacement?

Homework Equations

a is in m/s^2
v is in m/s
d is in m

The Attempt at a Solution

when t=0
displacement
at^2 = d = 0 = At^3 - Bt^4
velocity
at = v = 0 = At^2 - Bt^3

 

[Doc Al]

Please write the full expressions for v(t) and x(t).

 

[DrunkEngineer]

v(t) is velocity
dv/dt = At - Bt^2
dv = dt(At - Bt^2)

v(t) = \frac{At^{2}}{2} - \frac{Bt^{3}}{3} + Constant

x(t) is displacement
dx(t)/dt = (At^2)/2 - (Bt^3)/3 + Constant
dx(t) = ((At^2)/2 - (Bt^3)/3 + Constant)dt

x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + Constant

 

[Doc Al]

Good. What must each "Constant" equal?

 

[DrunkEngineer]

initial velocity
v(t) = \frac{At^{2}}{2} - \frac{Bt^{3}}{3} + v_{o}

initial displacement
x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + x_{o}

at t = 0;
For v(t):
v(0) = v_{o}

since v_{o} = 0 at time t = 0
v(0) = 0

For x(t):
x(0) = x_{o}

since x_{o} = 0 at time t = 0
x(0) = 0when :
x(t) = 0
x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + x_{o}

0 = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + 0

\frac{At^{3}}{6} = \frac{Bt^{4}}{12}

\frac{A}{6} = \frac{Bt}{12}

\frac{2A}{B} = t

 

[Doc Al]

Looks good to me.