AxeluteZero
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Position Function from Constant Acceleration Equation
The acceleration of a certain rocket is given by ax = bt, where b is a positive constant.
(a) Find the position function x(t) if x = x0 and v0 at t = 0. (Use x_0 for x0, v_0 for v0, b, and t as necessary.)
x(t) =
(Hint given): The velocity function is the time integral of the acceleration function. The position function is the time integral of the velocity function. The two integration constants can be determined by applying the given initial conditions when the time is equal to zero.
\Deltas = vi(\Delta) + (1/2)a(\Delta)2
ax = bt
so a = bt
v = int (a dt) = b * int (t dt) = b (.5t2)
x = b/2 int (t2) dt
Is my thinking in the right spot? I don't know if I'm not integrating correctly or if I need to plug the integrations into an equation...?
Homework Statement
The acceleration of a certain rocket is given by ax = bt, where b is a positive constant.
(a) Find the position function x(t) if x = x0 and v0 at t = 0. (Use x_0 for x0, v_0 for v0, b, and t as necessary.)
x(t) =
Homework Equations
(Hint given): The velocity function is the time integral of the acceleration function. The position function is the time integral of the velocity function. The two integration constants can be determined by applying the given initial conditions when the time is equal to zero.
\Deltas = vi(\Delta) + (1/2)a(\Delta)2
The Attempt at a Solution
ax = bt
so a = bt
v = int (a dt) = b * int (t dt) = b (.5t2)
x = b/2 int (t2) dt
Is my thinking in the right spot? I don't know if I'm not integrating correctly or if I need to plug the integrations into an equation...?
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