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Resposta:

Vamos dividir o percurso do motorista nos três trechos consecutivos [tex]\Delta x_1, \Delta x_2\,e\,\,\Delta x_3,[/tex] percorridos respectivamente às velocidades [tex]v_1, v_2\,\,e\,\,v_3,[/tex] nos respectivos intervalos de tempo [tex]\Delta t_1, \Delta t_2 \,e\,\,\Delta t_3.[/tex]

Trecho 1:

[tex]v_1 = \frac{\Delta x_1}{\Delta t_1}\\\\\Longleftrightarrow 40 = \frac{10}{\Delta t_1}\\\\\Longleftrightarrow \Delta t_1 = \frac{10}{40} = \frac{1}{4}\,\,h.[/tex]

Trecho 2:

[tex]v_2 = \frac{\Delta x_2}{\Delta t_2}\\\\\Longleftrightarrow 80 = \frac{10}{\Delta t_2}\\\\\Longleftrightarrow \Delta t_2 = \frac{10}{80} = \frac{1}{8}\,\,h.[/tex]

Trecho 3:

[tex]v_3 = \frac{\Delta x_3}{\Delta t_3}\\\\\Longleftrightarrow 30 = \frac{10}{\Delta t_3}\\\\\Longleftrightarrow \Delta t_3 = \frac{10}{30} = \frac{1}{3}\,\,h.[/tex]

Calculemos a velocidade média durante todo o percurso:

[tex]v_m = \frac{\Delta x_1 + \Delta x_2 + \Delta x_3}{\Delta t_1 + \Delta t_2 + \Delta t_3}\\\\\Longleftrightarrow v_m = \frac{10 + 10 + 10}{\frac{1}{4} + \frac{1}{8} + \frac{1}{3} }\\\\\Longleftrightarrow v_m = \frac{30}{\frac{17}{24} }\\\\\Longleftrightarrow \boxed{v_m = 42,4\,\,km/h.}[/tex]

Calculemos agora a média aritmética das velocidades:

[tex]\={v} = \frac{v_1 + v_2 + v_3}{3}\\\\\Longleftrightarrow \={v} = \frac{40 + 80 + 30}{3}\\\\\Longleftrightarrow \={v} = \frac{150}{3}\\\\\Longleftrightarrow \boxed{\={v} = 50\,\,km/h.}[/tex]

Perceba que, neste caso, [tex]v_m \neq \={v}.[/tex]