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3 years ago

5

A runner circles the track exactly 2 times for distance of 800m. it takes 4.0min. what is her average speed in m/s? what is her

average velocity?

1 answer:

Sindrei [870]3 years ago

5 0

If the runner ends where she started, then her total displacement is $\Delta x=0\,\mathrm m$ , which means her average velocity will be

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{0\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=0\,\dfrac{\mathrm m}{\mathrm s}$

The total distance she traversed, however, was $\Delta d=800\,\mathrm m$ , which gives her an average speed of

$\bar s=\dfrac{\Delta d}{\Delta t}=\dfrac{800\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=3.3\,\dfrac{\mathrm m}{\mathrm s}$

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

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Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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