Finding an Equation through Integrals

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SUMMARY

The discussion focuses on deriving the position function x(t) for a rocket with acceleration defined as ax = bt, where b is a constant. The solution involves integrating the acceleration to find the velocity function v(t) = 0.5 * b * t² + v0, and subsequently integrating the velocity to obtain the position function x(t). The integration constants are determined using initial conditions, specifically x(0) = x0 and v(0) = v0. The final position function is expressed as x(t) = (1/6) * b * t³ + v0 * t + x0.

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Position Function from Constant Acceleration 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.

[tex]\Delta[/tex]s = vi([tex]\Delta[/tex]) + (1/2)a([tex]\Delta[/tex])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...?
 
Last edited:
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You have to include a constant C when integrating. So v=0.5*b*t2+C.

If the initial velocity is v0, C=v0.

Now integrate v(t), do not forget the integration constant again.

ehild
 

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