Question

uma particula percorre em trajetória reta dois trechos. Em MN,manteve uma velocidade de 20km/h e no 2°trecho,NP,uma velocidade de 60km/h. Se o trecho NP é o dobro d trecho MN,determine a velocidade média no trecho MP.dê a resposta em km/h

Answer 1

Trecho MN:

$v_{1}=20~km/h\\\Delta S_{1}=x\\\Delta t_{1}=~?$
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$v_{1}=\dfrac{\Delta S_{1}}{\Delta t_{1}}$

Isolando Δt₁:

$\Delta t_{1}=\dfrac{\Delta S_{1}}{v_{1}}\\\\\boxed{\boxed{\Delta t_{1}=\frac{x}{20}~h}}$
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Trecho NP:

$v_{2}=60~km/h\\\Delta S_{2}=2x~~~(dobro~de~MN)\\\Delta t_{2}=~?$
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$v_{2}=\dfrac{\Delta S_{2}}{\Delta t_{2}}$

Isolando Δt₂:

$\Delta t_{2}=\dfrac{\Delta S_{2}}{v_{2}}\\\\\\\Delta t_{2}=\dfrac{2x}{60}\\\\\\\boxed{\boxed{\Delta t_{2}=\dfrac{x}{30}~h}}$
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Finalmente:

$v_{m}=?\\\Delta S=\Delta S_{1}+\Delta S_{2}=x+2x=3x~km\\\\\Delta t=\Delta t_{1}+\Delta t_{2}=\frac{x}{20}+\frac{x}{30}=\frac{30x+20x}{20*30}=\frac{x}{12}~h$

Achando a velocidade média no percurso:

$v_{m}=\dfrac{\Delta S}{\Delta t}\\\\\\v_{m}=\dfrac{3x}{(\frac{x}{12})}\\\\\\v_{m}=3x*\dfrac{12}{x}\\\\\\v_{m}=3*12\\\\\boxed{\boxed{v_{m}=36~km/h}}$