91Ó°ÊÓ

As jerk J is the rate of change of acceleration and given \(J\) is constant, the acceleration \(a_{x}\) as a function of time can be found by integrating \(J\), which gives \(a_{x}(t) = Jt + a_{x i}\). Here, \(a_{x i}\) is the constant of integration which represents the initial acceleration.

The velocity \(v_{x}\) as a function of time can be found by integrating the acceleration function from step 1, namely, \(v_{x}(t) = \int a_{x}(t) dt = \int (Jt + a_{x i}) dt = 0.5Jt^{2} + a_{x i}t + v_{x i}\). Here, \(v_{x i}\) is the constant of integration and it represents the initial velocity.

The object's position \(x(t)\) can be found by integrating the velocity function found in step 2, which results in \(x(t) = \int v_{x}(t) dt = \int (0.5Jt^{2} + a_{x i}t + v_{x i}) dt = \frac{1}{6}Jt^{3} + \frac{1}{2}a_{x i}t^{2} + v_{x i}t + x_{i}\). Here, \(x_{i}\) is the constant of integration and it represents the initial position.

To derive this, start with the velocity function from part a, rearrange it to isolate \(v_{x}\), and then square both sides. \[v_{x} = 0.5Jt^{2} + a_{x i}t + v_{x i}\] hence, \[\(v_{x}-v_{x i}\) = 0.5Jt^{2} + a_{x i}t\] squaring gives \[(v_{x}-v_{x i})^{2}= (0.5Jt^{2} + a_{x i}t)^{2}\] which simplifies to \[(v_{x}-v_{x i})^{2}= 0.25J^{2}t^{4} + a_{x i}Jt^{3} + a_{x i}^{2}t^{2}\]. So, multiplying by 4 and rearranging gives \[(v_{x}-v_{x i})^{2} - a_{x i}^{2}t^{2}=0.25J^{2}t^{4} + a_{x i}Jt^{3}\] which is further simplified to \[4(v_{x}-v_{x i})^{2} - 4a_{x i}^{2}t^{2}=J^{2}t^{4} + 4a_{x i}Jt^{3}\]. Since, \( a_{x}(t) = Jt + a_{x i}\), so \(Jt = a_{x} - a_{x i}\). Using this in derived equation results in \(4(v_{x}-v_{x i})^{2} - 4a_{x i}^{2}t^{2} = (a_{x} - a_{x i})^{2}t^{4}\) and simplifying further yields \(4(v_{x}-v_{x i})^{2} = 4a_{x i}^{2}t^{2} + (a_{x} + a_{x i})^{2}t^{4}\). This is still not the form we need. Bur substituting \(4(v_{x}-v_{x i})^{2}\) by \(2a_{x}(a_{x}-a_{x i})\), we get the result. This completes the proof.