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

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The discussion focuses on solving for the time t when a body, initially at zero displacement and velocity, returns to zero displacement under the acceleration equation a = At - Bt². The derived expressions for velocity and displacement are v(t) = (At²/2) - (Bt³/3) and x(t) = (At³/6) - (Bt⁴/12), respectively. By substituting the conditions at t = 0, both initial velocity and displacement constants are confirmed to be zero. Setting the displacement equation to zero leads to the relationship (2A/B) = t, indicating the next time the body has zero displacement. The final result shows that t can be calculated directly from the constants A and B.
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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
 
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Please write the full expressions for v(t) and x(t).
 
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
 
Good. What must each "Constant" equal?
 
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
 
Looks good to me.
 
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