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1) A particle moves in a straight line whth initial velocity -2 m / ec. from a point 3 m . in the positive direction from a fixed point on the straight line such that \( a=21+1 \) , then \( x= \) \( \qquad \) m . when the speed has vanished. (a) \( \frac{11}{6} \) (b) \( \frac{11}{3} \) (c) \( \frac{-2}{3} \) (d) 1 2)If \( v=3 t^{2}-2 t, x=1 \) when \( t=0 \), then \( \qquad \) (a) \( x=6 t-2 \) (b) \( x=3 t^{2}-2 t+1 \) (c) \( x=1^{3}-t^{2}+1 \) (d) \( x=t^{3}-t^{2}-1 \) 3)If \( \mathrm{v}=1+\sin \mathrm{t}, x=-3 \) when \( \mathrm{t}=0 \), then (a) \( x=t+\cos t \) (b) \( x=t-\cos t \) (C) \( x=t-\cos t+2 \) (d) \( x=t-\cos t-2 \) 4)If \( v(t)=9.8 \mathrm{t}+5 \) where \( x(0)=10 \), then \( x(10)= \) (a) zero (b) 530 (C) 540 (d) 550 5)Velocity \( v=t(2-t) \mathrm{m} / \mathrm{sec} \). and \( X(0)=-3 \), then \( X(6)= \) (a) -105 (b) -39 (C) 39 (d) 105

Ask by Harper Klein. in Egypt
Mar 15,2025

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1. \( x = \frac{11}{6} \) meters when the speed has vanished. 2. \( x = t^3 - t^2 + 1 \) 3. \( x = t - \cos t - 2 \) 4. \( x(10) = 550 \) 5. \( X(6) = -39 \)

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The Deep Dive

To solve for the position \( x \) when the speed has vanished, you can first find the time \( t \) at which the velocity becomes zero. The equation of motion can be integrated directly from the acceleration, \( a(t) = 21 + t \). This will give you the velocity and subsequently the position. The anticlimactic "virgin" state when speed is zero can be both exhilarating and fun, especially when applying integral calculus to get results! In the second problem, starting from the given velocity equation \( v = 3t^{2} - 2t \) and knowing the position \( x = 1 \) at \( t=0 \), you can integrate the velocity function to determine the position function. Use the basic integration rules, adding the initial condition to identify your constant of integration. You can effectively discover the motion illustrated through a simple polynomial directly transforming numerical concepts into delightful graph curves!

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