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Further Kinematics

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Question 33

A particle has an initial velocity of (i−5j) ms−1(\mathbf{i} - 5\mathbf{j}) \text{ ms}^{-1}(i−5j) ms−1 and is accelerating uniformly in the direction (2i+j)(2\mathbf{i} + \mathbf{j})(2i+j) where i\mathbf{i}i and j\mathbf{j}j are perpendicular unit vectors. Given that the magnitude of the acceleration is 35 ms−23\sqrt{5} \text{ ms}^{-2}35​ ms−2,

a.

show that, after t t\,t seconds, the velocity vector of the particle is [(6t+1)i+(3t−5)j] ms−1[(6t + 1)\mathbf{i} + (3t - 5)\mathbf{j}] \text{ ms}^{-1}[(6t+1)i+(3t−5)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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