Question

Compute your average velocity in the following two cases: (a) You walk 88.0 m at a speed of 2.24 m/s and then run 88.0 m at a speed of 3.28 m/s along a straight track. (b) You walk for 1.00 min at a speed of 2.24 m/s and then run for 1.50 min at 3.28 m/s along a straight track.

Answer 1

Answer on Question #51359, Physics, Mechanics | Kinematics | Dynamics

Compute your average velocity in the following two cases: (a) You walk 88.0 m at a speed of 2.24 m/s and then run 88.0 m at a speed of 3.28 m/s along a straight track. (b) You walk for 1.00 min at a speed of 2.24 m/s and then run for 1.50 min at 3.28 m/s along a straight track.

Solution.

a) From the formula $\nu = \frac{s}{t}$ we have

$$2, 2 4 = \frac {8 8}{t _ {1}} \Rightarrow t _ {1} = \frac {8 8}{2 , 2 4} = \frac {1 1}{0 , 2 8};$$ $$3, 2 8 = \frac {8 8}{t _ {2}} \Rightarrow t _ {2} = \frac {8 8}{3 , 2 8} = \frac {1 1}{0 , 4 1}.$$

Thus, whole distance $s = 88 + 88 = 176m$ has been passed for

$$\begin{array}{l} t = t _ {1} + t _ {2} = \frac {1 1}{0 , 2 8} + \frac {1 1}{0 , 4 1} = \frac {1 1 0 0}{2 8} + \frac {1 1 0 0}{4 1} = 1 1 0 0 \cdot \frac {4 1 + 2 8}{1 1 4 8} = \\ = 1 1 0 0 \cdot \frac {6 9}{1 1 4 8} = 2 7 5 \cdot \frac {6 9}{2 8 7} = \frac {1 8 9 7 5}{2 8 7}. \\ \end{array}$$

So

$$\nu = \frac {s}{t} = \frac {1 7 6}{\frac {1 8 9 7 5}{2 8 7}} = \frac {5 0 5 1 2}{1 8 9 7 5} \approx 2, 6 6 m / s$$

b) We know that

$$\begin{array}{l} 1 \min = 6 0 s; \\ 1, 5 0 \min = 9 0 s. \\ \end{array}$$

From the formula $\nu = \frac{s}{t}$ we have

$$\begin{array}{l} 2, 2 4 = \frac {s _ {1}}{6 0} \Rightarrow s _ {1} = 6 0 \cdot 2, 2 4 = 1 3 4, 4; \\ 3, 2 8 = \frac {s _ {2}}{9 0} \Rightarrow s _ {2} = 9 0 \cdot 3, 2 8 = 2 9 5, 2. \\ \end{array}$$

Thus, whole distance $s = 134,4 + 295,2 = 429,6m$ has been passed for $t = t_1 + t_2 = 60 + 90 = 150s$ .

So

$$\nu = \frac {s}{t} = \frac {429,6}{150} = 2,864\,m/s.$$

Answer: a) $\nu \approx 2,66\,m/s$ ; b) $\nu = 2,864\,m/s$.

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