Os freios de um carro são capazes de produzir uma desaceleração de 5,2m/s2
.
a. Se você está dirigindo a 140km/h e avista, de repente, um posto policial, qual o tempo mínimo
necessário para reduzir a velocidade até o limite permitido de 100km/h ?
b. No caso de acelerar a 2m/s2 depois do posto em quanto tempo atinge 120km/h?
Soluções para a tarefa
Respondido por
1
a)
![v=100~km/h=(100/3,6)~m/s=(1000/36)~m/s=(250/9)~m/s\\v_{0}=140~km/h=(140/3,6)~m/s=(1400/36)~m/s=(350/9)~m/s\\a=-5,2~m/s^{2}\\t=? v=100~km/h=(100/3,6)~m/s=(1000/36)~m/s=(250/9)~m/s\\v_{0}=140~km/h=(140/3,6)~m/s=(1400/36)~m/s=(350/9)~m/s\\a=-5,2~m/s^{2}\\t=?](https://tex.z-dn.net/?f=v%3D100%7Ekm%2Fh%3D%28100%2F3%2C6%29%7Em%2Fs%3D%281000%2F36%29%7Em%2Fs%3D%28250%2F9%29%7Em%2Fs%5C%5Cv_%7B0%7D%3D140%7Ekm%2Fh%3D%28140%2F3%2C6%29%7Em%2Fs%3D%281400%2F36%29%7Em%2Fs%3D%28350%2F9%29%7Em%2Fs%5C%5Ca%3D-5%2C2%7Em%2Fs%5E%7B2%7D%5C%5Ct%3D%3F)
Equação horária da velocidade:
![\frac{250}{9}=\frac{350}{9}-5,2t\\\\\frac{250}{9}-\frac{350}{9}=-5,2t\\\\-5,2t=\frac{100}{9}\\\\-5,2t\approx11,1\\\\t\approx-11,1/5,2\\\\\boxed{\boxed{t\approx2,13~s}} \frac{250}{9}=\frac{350}{9}-5,2t\\\\\frac{250}{9}-\frac{350}{9}=-5,2t\\\\-5,2t=\frac{100}{9}\\\\-5,2t\approx11,1\\\\t\approx-11,1/5,2\\\\\boxed{\boxed{t\approx2,13~s}}](https://tex.z-dn.net/?f=%5Cfrac%7B250%7D%7B9%7D%3D%5Cfrac%7B350%7D%7B9%7D-5%2C2t%5C%5C%5C%5C%5Cfrac%7B250%7D%7B9%7D-%5Cfrac%7B350%7D%7B9%7D%3D-5%2C2t%5C%5C%5C%5C-5%2C2t%3D%5Cfrac%7B100%7D%7B9%7D%5C%5C%5C%5C-5%2C2t%5Capprox11%2C1%5C%5C%5C%5Ct%5Capprox-11%2C1%2F5%2C2%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Bt%5Capprox2%2C13%7Es%7D%7D)
b)
![v=120~km/h=(120/3,6)~m/s=(1200/36)=(300/9)~m/s\\v_{0}=100~km/h=(250/9)~m/s\\a=+2~m/s^{2}\\t=? v=120~km/h=(120/3,6)~m/s=(1200/36)=(300/9)~m/s\\v_{0}=100~km/h=(250/9)~m/s\\a=+2~m/s^{2}\\t=?](https://tex.z-dn.net/?f=v%3D120%7Ekm%2Fh%3D%28120%2F3%2C6%29%7Em%2Fs%3D%281200%2F36%29%3D%28300%2F9%29%7Em%2Fs%5C%5Cv_%7B0%7D%3D100%7Ekm%2Fh%3D%28250%2F9%29%7Em%2Fs%5C%5Ca%3D%2B2%7Em%2Fs%5E%7B2%7D%5C%5Ct%3D%3F)
Equação horária da velocidade:
![\frac{300}{9}=\frac{250}{9}+2\cdot t\\\\\frac{300-250}{9}=2t\\\\\frac{50}{9}=2t\\\\\frac{25}{9}=t\\\\\\\boxed{\boxed{t\approx2,78~s}} \frac{300}{9}=\frac{250}{9}+2\cdot t\\\\\frac{300-250}{9}=2t\\\\\frac{50}{9}=2t\\\\\frac{25}{9}=t\\\\\\\boxed{\boxed{t\approx2,78~s}}](https://tex.z-dn.net/?f=%5Cfrac%7B300%7D%7B9%7D%3D%5Cfrac%7B250%7D%7B9%7D%2B2%5Ccdot+t%5C%5C%5C%5C%5Cfrac%7B300-250%7D%7B9%7D%3D2t%5C%5C%5C%5C%5Cfrac%7B50%7D%7B9%7D%3D2t%5C%5C%5C%5C%5Cfrac%7B25%7D%7B9%7D%3Dt%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7Bt%5Capprox2%2C78%7Es%7D%7D)
Equação horária da velocidade:
b)
Equação horária da velocidade:
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