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Ad libitum [116K]

3 years ago

13

A car traveling at 60 km/h will skid 30 m when its brakes are locked. If the same car is traveling at 180 km/h, what will be the

skidding distance? Be sure to show all work to support your answer.

1 answer:

cestrela7 [59]3 years ago

8 0

Converting 60 and 180 kmph into m/s
that's 16.67 m/s and 50 m/s
now
16.67*16.67 = 2*a*30
277.88/60 = a
4.631 m/s^2
as the vehicle is same it will decelerate with the same value
and thus
50*50 = 2*4.631*s
2500/9.262 = s
that's around 270 m

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• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

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• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

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$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

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$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

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