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A particle is at the position with Cartesian components $(5.0 m, 4.0 m)$ and $5.0 s$ later it is at the position with components $(8.0 m, 0.0 m)$. What is the magnitude of its average velocity?

I know the answer is $1.0 m/s$.

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1 Answer 1

Average velocity is defined as the change in position divided by the change in time. $$ \frac{\Delta x}{\Delta t} $$ In our case, we are given $\Delta t = 5 \textrm{ s}$. We calculate $\Delta x$ by subtracting the initial position from the final position. $$ x - x_0 = (8, 0) - (5, 4) = (3,-4) \textrm{ m} $$ Doing the divison, we have $$ \text{average velocity } = \left(\frac{3}{5}, -\frac{4}{5}\right) \textrm{ m}/\textrm{s}. $$ We calculate the magnitude using the formula for magnitude $$ |{\text{average velocity}}|=\sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2} = 1 \textrm{ m}/\textrm{s} $$

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