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Further Kinematics

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Question 22

A small object starts with an initial velocity u=(−8i+2j) ms−1\mathbf{u} = (-8\mathbf{i} + 2\mathbf{j}) \text{ ms}^{-1}u=(−8i+2j) ms−1. It accelerates at a constant rate in the direction (i+j)(\mathbf{i} + \mathbf{j})(i+j). Given that the magnitude of the acceleration is 52 ms−25\sqrt{2} \text{ ms}^{-2}52​ ms−2,

a.

Show that the velocity vector v\mathbf{v}v after t t\,t seconds is given by v=[(5t−8)i+(5t+2)j] ms−1\mathbf{v} = [(5t - 8)\mathbf{i} + (5t + 2)\mathbf{j}] \text{ ms}^{-1}v=[(5t−8)i+(5t+2)j] ms−1.

[6]
b.

Using your answer to part (a), or otherwise, find the value of t t\,t for which the speed of the particle is at its minimum.

[5]

Further Kinematics Questions

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