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Suppose an airplane is acted on by three forces: gravity, wind and engine thrust. Assume that the force vector for gravity is mg=m0,0,32m \mathbf{g}=m\langle 0,0,-32\rangle, the force vector for wind is w=0,1,0\mathbf{w}=\langle 0,1,0\rangle for 0t10 \leq t \leq 1 and w=0,2,0\mathbf{w}=\langle 0,2,0\rangle for t>1, and the force vector for engine thrust is e=2t,0,24\mathbf{e}=\langle 2 t, 0,24\rangle. Newton's second law of motion gives us ma=mg+w+e. Assume that m=1 and the initial velocity vector is v(0)=100,0,10\mathbf{v}(0)=\langle 100,0,10\rangle. Show that the velocity vector for 0t1 is v(t)=t2+100,t,108t0 \leq t \leq 1 \text { is } \mathbf{v}(t)=\left\langle t^{2}+100, t, 10-8 t\right\rangle. For t>1, integrate the equation a=g+w+e, to get v(t)=t2+a,2t+b,8t+c\mathbf{v}(t)=\left\langle t^{2}+a, 2 t+b,-8 t+c\right\rangle, for constants a, b and c. Explain (on physical grounds) why the function v(t) should be continuous and find the values of the constants that make it so. Show that v(t) is not differentiable. Given the nature of the force function, why does this make sense?


Answered 1 year ago
Answered 1 year ago
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This problem tries to asks us the concept of differentiability, and how piecewise functions in times give out a function that cannot be differentiable. This case is one of them.

We are given with force vectors for gravity mg=0,0,32m\textbf{g}=\langle0,0,-32\rangle, for wind w=0,1,0\textbf{w}=\langle0,1,0\rangle for 0t10\leq t\leq 1 and w=0,2,0\textbf{w}=\langle0,2,0\rangle for t>1t>1, and for the engine e=2t,0,24\textbf{e}=\langle2t,0,24\rangle.

We also assume that m=1m=1. Initial velocity (v(0))(\textbf{v}(0)) is 100,0,10\langle100,0,10\rangle. By Newton's 2nd law of motion, ma=mg+w+em\textbf{a}=m\textbf{g}+\textbf{w}+\textbf{e}, but since m=1m=1, our final vector function for acceleration is equal to a=g+w+e\textbf{a}=\textbf{g}+\textbf{w}+\textbf{e}

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