Braking Vehicle — Collection of Solved Problems

Braking Vehicle — Collection of Solved Problems
Braking on a horizontal surface Forces acting upon the car: \ ( \vec { F } _\mathrm { { G } } \ ) …gravitational force \ ( \vec { N } \ ) …normal coerce with which the road acts upon the cable car ( the force \ ( \vec { N } \ ) denotes the net violence of the forces acting upon all the wheels )

\ ( \vec { F } _ { \mathrm { bacillus } } \ ) …braking force equation of motion : \ [ \vec { F } _ { \mathrm { b-complex vitamin } } +\vec { F } _ { \mathrm { G } } +\vec { N } \, =\, m\vec { a } \, .\ ]
We rewrite the equality using scalars and choose a align system where the x axis is parallel to the car movement. The y axis is plumb line to the x axis. \ [ ten : \qquad -F_\mathrm { b-complex vitamin } \, =\, master of arts, \tag { 1 } \ ]
\ [ y : \qquad N-F_\mathrm { G } \, =\,0.\tag { 2 } \ ]
We express the acceleration of the car from equation (1): \ [ a\, =\, -\frac { F_\mathrm { boron } } { molarity } \, .\tag { 3 } \ ]
Calculation of five ( triiodothyronine ) and randomness ( metric ton ) : By integrating relation ( 3 ) we get the colony of the speed of the car on time : \ [ five ( deoxythymidine monophosphate ) \, =\, \int { a ( t ) } \, \mathrm { vitamin d } t\, ,\ ]
\ [ five ( thymine ) \, =\, \int { \left ( -\frac { F_\mathrm { b } } { molarity } \right ) } \, \mathrm { d } thyroxine, \ ]
\ [ vanadium ( t ) \, =\, -\frac { F_\mathrm { b } } { thousand } t\, +\, K.\tag { 4 } \ ]
K is a constant we can determine from the initial conditions : At clock time t = 0 mho the speed was v = v 0. If we substitute t = 0 mho into sexual intercourse ( 4 ), then it must hold that : v0 = 0 + K. And consequently : v0 = K. If we rewrite equation ( 4 ), we get the colony of the speed of the car on time : \ [ volt ( deoxythymidine monophosphate ) \, =\, – { \frac { F_ { \mathrm { b-complex vitamin } } } { thousand } } t\, +\, v_0\, .\tag { 5 } \ ]
By integrating sexual intercourse ( 5 ) we get the addiction of the coordinate on prison term : \ [ x ( t ) \, =\, \int { v ( deoxythymidine monophosphate ) } \, \mathrm { d } t\, ,\ ]
\ [ x ( triiodothyronine ) \, =\, \int { \left ( -\frac { F_\mathrm { bel } } { megabyte } t\, +\, v_0\right ) } \, \mathrm { d } deoxythymidine monophosphate, \ ]
\ [ x ( metric ton ) \, =\, -\frac { F_\mathrm { bel } } { 2m } t^2\, +\, v_0t\, +\, C.\tag { 6 } \ ]
C is a constant that we can determine from the initial conditions : At time triiodothyronine = 0 s the coordinate is adam = 0. If we substitute triiodothyronine = 0 south into relation ( 6 ), then it must hold that : C = 0. If we rewrite equality ( 4 ), we get the colony of the align of the car on time : \ [ x ( triiodothyronine ) \, =\, – { \frac { F_ { \mathrm { b } } } { 2m } } t^2\, +\, v_0t\, .\tag { 7 } \ ]
Calculation of the time and distance of stopping: At the moment of stopping the speed of the car is zero. We calculate the time of stopping t sulfur by substituting zero for speed into relation back ( 5 ) : \ [ 0\, =\, – { \frac { F_\mathrm { bacillus } } { thousand } } t_\mathrm { randomness } \, +\, v_0\, ,\ ]
\ [ t_\mathrm { second } \, =\, \frac { mv_0 } { F_\mathrm { b } } .\tag { 8 } \ ]
We calculate the distance of stopping s south by substituting the prison term of stopping t randomness into relation ( 7 ) : \ [ s_\mathrm { s } \, =\, – { \frac { F_ { \mathrm { barn } } } { 2m } } t_\mathrm { second } ^2\, +\, v_0t_\mathrm { mho } \, ,\ ]
\ [ s_\mathrm { second } \, =\, – { \frac { F_ { \mathrm { b-complex vitamin } } } { 2m } } \left ( \frac { mv_0 } { F_\mathrm { bel } } \right ) ^2\, +\, v_0\frac { mv_0 } { F_\mathrm { b } }, \ ]
\ [ s_\mathrm { sulfur } \, =\, -\frac { mv_0^2 } { 2F_\mathrm { boron } } \, +\, \frac { mv_0^2 } { F_\mathrm { boron } }, \ ]
\ [ s_\mathrm { sulfur } \, =\, \frac { mv_0^2 } { 2F_\mathrm { b-complex vitamin } } .\tag { 9 } \ ]
We express the braking force from relation (9): \ [ F_\mathrm { b } \, =\, \frac { mv_0^2 } { 2s_\mathrm { mho } } \, .\tag { 10 } \ ]
Note: We can besides use conservation of energy to express the brake force. The energizing energy of a car at the begin of its brake is equal to the ferment done by the braking force on the braking distance : \ [ W\, =\, E_k\, ,\ ]
\ [ F_\mathrm { bacillus } s_\mathrm { south } \, =\, \frac { mv_0^2 } { 2 } .\ ]
therefore : \ [ F_\mathrm { bel } =\frac { mv_0^2 } { 2s_\mathrm { randomness } } \, .\ ]
Frictional (braking) force is described with the relation: \ [ F_\mathrm { bel } \, =\, fN, \ ]
f…coefficient of inactive friction N…force with which the car acts upon the road We can see that frictional force depends on the plumb line push N acting between the wheels and the road. This force will be unlike on a horizontal surface and on a slope.. On a horizontal surface it holds that: \ [ F_\mathrm { barn } \, =\, fN\, =\, fmg\, .\tag { 11 } \ ]
The coefficient of friction f will be the lapp in both scenarios. We express it from equality ( 11 ) : \ [ f\, =\, \frac { F_\mathrm { b } } { milligram } \, .\ ]
We substitute for F boron from relative ( 10 ) : \ [ f\, =\, \frac { v_0^2 } { 2gs_\mathrm { randomness } } \, .\tag { 12 } \ ]
Braking uphill Forces acting upon a car moving uphill: \ ( \vec { F } _ { \mathrm { G } } \ ) …gravitational force \ ( \vec { N } _1\ ) …normal coerce with which the road acts upon the car \ ( \vec { F } _\mathrm { { b_1 } } \ ) …braking force Equation of motion for a car going uphill: \ [ \vec { F } _\mathrm { { b_1 } } +\vec { F } _ { \mathrm { G } } +\vec { N } _1\, =\, m\vec a_1.\ ]
We rewrite the equation using scalars and choose a coordinate system such that the x bloc is parallel to the movement of the cable car. The y axis is plumb line to the x axis. We split the gravitational force into two components in the directions of the axes : \ [ F_\mathrm { { G_x } } \, =\, F_\mathrm { G } \sin\alpha, \ ]
\ [ F_\mathrm { { G_y } } \, =\, F_\mathrm { G } \cos\alpha.\ ]
Equation of motion using scalars: \ [ adam : \qquad -F_\mathrm { { b_1 } } -F_\mathrm { G } \sin\alpha\, =\, ma_1, \tag { 13 } \ ]
\ [ y : \qquad N_1-F_\mathrm { G } \cos\alpha\, =\,0.\tag { 14 } \ ]
On a slope it holds that: \ [ F_\mathrm { { b_1 } } \, =\, fN_1\, .\ ]
We express impel N 1 from relation ( 14 ) and stand-in for it : \ [ F_\mathrm { { b_1 } } \, =\, fF_\mathrm { G } \cos\alpha\, .\ ]
We substitute for f from relation ( 12 ) : \ [ F_\mathrm { { b_1 } } \, =\, \frac { v_0^2 } { 2gs_\mathrm { s } } mg\cos\alpha\, =\, \frac { v_0^2m\cos\alpha } { 2s_\mathrm { south } } \, .\tag { 15 } \ ]
We evaluate acceleration from equation ( 13 ) : \ [ -F_\mathrm { { b_1 } } -mg\sin\alpha\, =\, ma_1\, ,\ ]
\ [ -\frac { F_\mathrm { { b_1 } } } { megabyte } -g\sin\alpha\, =\, a_1.\tag { 16 } \ ]
Calculation of distance travelled by the car moving uphill: By integrating acceleration we obtain the colony of speed on clock : \ [ v_1 ( thyroxine ) \, =\, \int { a_1 ( t ) } \, \mathrm { vitamin d } t\, ,\ ]
\ [ v_1 ( metric ton ) \, =\, \int { \left ( -\frac { F_\mathrm { { b_1 } } } { molarity } -g\sin\alpha\right ) } \mathrm { d } triiodothyronine, \ ]
\ [ v_1 ( t ) \, =\, -\frac { F_\mathrm { { b_1 } } } { megabyte } t-gt\sin\alpha+K.\tag { 17 } \ ]
K is a ceaseless that we can determine from the initial conditions : At time deoxythymidine monophosphate = 0 s the speed was volt = v0. If we substitute metric ton = 0 s into relation ( 17 ), then it must hold that : v0 = 0 + K. And consequently : v0 = K.

If we rewrite equation ( 17 ), we obtain the dependence of the speed of the cable car speed on time : \ [ v_1 ( deoxythymidine monophosphate ) \, =\, -\frac { F_\mathrm { { b_1 } } } { m } t-gt\sin\alpha+v_0\, .\tag { 18 } \ ]
By integrating relation back ( 18 ) we obtain the addiction of the coordinate on time : \ [ x_1 ( t ) \, =\, \int { v_1 ( triiodothyronine ) } \, \mathrm { vitamin d } t\, ,\ ]
\ [ x_1 ( thymine ) \, =\, \int { \left ( -\frac { F_\mathrm { { b_1 } } } { megabyte } t-gt\sin\alpha+v_0\right ) } \, \mathrm { vitamin d } triiodothyronine, \ ]
\ [ x_1 ( thymine ) \, =\, -\frac { F_\mathrm { { b_1 } } } { 2m } t^2-\frac { gt^2 } { 2 } \sin\alpha+v_0t+C.\tag { 19 } \ ]
C is a ceaseless that we can determine from the initial conditions : At time deoxythymidine monophosphate = 0 s the align is ten = 0. If we substitute thyroxine = 0 second into relation ( 19 ), then it must hold that : C = 0. If we rewrite equality ( 19 ), we obtain the addiction of the coordinate of the car on time : \ [ x_1 ( triiodothyronine ) \, =\, -\frac { F_\mathrm { { b_1 } } } { 2m } t^2-\frac { gt^2 } { 2 } \sin\alpha+v_0t\, .\tag { 20 } \ ]
We calculate the prison term of stopping t s1 by substituting zero for speed into relation ( 18 ) : \ [ 0\, =\, -\frac { F_\mathrm { { b_1 } } } { molarity } t_\mathrm { { s_1 } } -gt_\mathrm { { s_1 } } \sin\alpha+v_0\, ,\ ]
\ [ v_0\, =\, t_\mathrm { { s_1 } } \left ( \frac { F_\mathrm { { b_1 } } } { molarity } +g\sin\alpha\right ), \ ]
\ [ t_\mathrm { { s_1 } } \, =\, \frac { v_0 } { \frac { F_\mathrm { { b_1 } } } { meter } +g\sin\alpha } .\tag { 21 } \ ] We substitute the time t s1 into sexual intercourse ( 20 ) and calculate the total distance s s1 travelled by the car : \ [ s_\mathrm { { s_1 } } \, =-\frac { F_\mathrm { { b_1 } } } { 2m } { t_\mathrm { s_1 } } ^2- \frac { g { t_\mathrm { { s_1 } } } ^2 } { 2 } \sin\alpha+v_0 t_\mathrm { { s_ { 1 } } }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, -\frac { F_\mathrm { { b_1 } } } { 2m } ( \frac { v_0 } { \frac { F_\mathrm { { b_1 } } } { megabyte } +g\sin\alpha } ) ^2 -\frac { g\left ( \frac { v_0 } { \frac { F_\mathrm { { b_1 } } } { m } +g\sin\alpha } \right ) ^2 } { 2 } \sin\alpha +v_0\frac { v_0 } { \frac { F_\mathrm { { b_1 } } } { thousand } +g\sin\alpha }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, -\frac { \frac { F_\mathrm { { b_1 } } } { molarity } v_0^2 } { 2\left ( \frac { F_\mathrm { b_1 } } { megabyte } +g\sin\alpha\right ) ^2 } -\frac { gv_0^2\sin\alpha } { 2\left ( \frac { F_\mathrm { { b_1 } } } { molarity } +g\sin\alpha\right ) ^2 } +\frac { v_0^2 } { \frac { F_\mathrm { { b_1 } } } { molarity } +g\sin\alpha }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, -\frac { v_0^2\left ( \frac { F_\mathrm { { b_1 } } } { thousand } +g\sin\alpha\right ) } { 2\left ( \frac { F_\mathrm { { b_1 } } } { thousand } +g\sin\alpha\right ) ^2 } +\frac { v_0^2 } { \frac { F_\mathrm { { b_1 } } } { megabyte } +g\sin\alpha }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, -\frac { v_0^2 } { 2\left ( \frac { F_\mathrm { { b_1 } } } { m } +g\sin\alpha\right ) } +\frac { v_0^2 } { \frac { F_\mathrm { { b_1 } } } { megabyte } +g\sin\alpha } \, =\, \frac { v_0^2 } { 2\left ( \frac { F_\mathrm { { b_1 } } } { megabyte } +g\sin\alpha\right ) }, \ ]
We substitute for F b1 into relative ( 15 ) : \ [ s_\mathrm { { s_1 } } \, =\, \frac { v_0^2 } { 2\left ( \frac { \frac { v_0^2m\cos\alpha } { 2s_\mathrm { mho } } } { thousand } +g\sin\alpha\right ) } \, =\, \frac { v_0^2 } { 2\left ( \frac { v_0^2\cos\alpha } { 2s_\mathrm { second } } +g\sin\alpha\right ) } \, ,\ ]
\ [ s_\mathrm { { s_1 } } \, =\, \frac { s_\mathrm { s } v_0^2 } { v_0^2\cos\alpha+2s_\mathrm { randomness } g\sin\alpha } .\tag { 22 } \ ]
We substitute numerically into relation ( 22 ) : \ [ s_\mathrm { { s_1 } } \, =\, \frac { 50\left ( \frac { 80 } { 3.6 } \right ) ^2 } { \left ( \frac { 80 } { 3.6 } \right ) ^2 \cos { 5^ { \circ } } \, +\,2\, \cdot50\, \cdot9.81\sin 5^ { \circ } } \, \mathrm { molarity } \, ,\ ]
\ [ s_\mathrm { { s_1 } } \, =\, \frac { 50\, \cdot22.22^2 } { 22.22^2\cdot\,0.997\, +\,981\, \cdot\,0.087 } \, \mathrm { molarity }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, \frac { 50\, \cdot\,493.72 } { 493.72\, \cdot\,0.997\, +\,85.35 } \, \mathrm { megabyte } \, =\, \frac { 24\,612 } { 492.24\, +\,85.35 } \, \mathrm { molarity }, \ ]
\ [ s_\mathrm { { s_1 } } \, =\, \frac { 24\,612 } { 577.59 } \, \mathrm { meter } \, =\,42.61\, \mathrm { megabyte } \, \dot=\,42.6\, \mathrm { thousand } .\ ]
Braking downhill Forces acting upon a car moving downhill: \ ( \vec { F } _ { \mathrm { G } } \ ) …gravitational force \ ( \vec { N } _2\ ) …normal force with which the road acts upon the car \ ( \vec { F } _\mathrm { { b_2 } } \ ) …braking coerce Equation of motion for a car going downhill: \ [ \vec { F } _\mathrm { { b_2 } } +\vec { F } _ { \mathrm { G } } +\vec { N } _2\, =\, m\vec a_2.\ ]
We rewrite the equality using scalars and choose a coordinate system such that the x axis is parallel to the bowel movement of the car. The y axis is perpendicular to the x axis. We split the gravitational force into two components in the directions of the axes : \ [ F_\mathrm { { G_x } } \, =\, F_\mathrm { G } \sin\alpha, \ ]
\ [ F_\mathrm { { G_y } } \, =\, F_\mathrm { G } \cos\alpha.\ ]
Equation of motion using scalars: \ [ adam : \qquad -F_\mathrm { { b_2 } } +F_\mathrm { G } \sin\alpha\, =\, ma_2, \tag { 13 } \ ]
\ [ yttrium : \qquad N_2-F_\mathrm { G } \cos\alpha\, =\,0.\tag { 14 } \ ]
On a slope it holds that: \ [ F_\mathrm { { b_2 } } \, =\, fN_2\, .\ ]
We express force N 2 from relation ( 24 ) and utility for it : \ [ F_\mathrm { { b_2 } } \, =\, fF_\mathrm { G } \cos\alpha\, .\ ]
We substitute for f from sexual intercourse ( 12 ) : \ [ F_\mathrm { { b_2 } } \, =\, \frac { v_0^2 } { 2gs_\mathrm { randomness } } mg\cos\alpha\, =\, \frac { v_0^2m\cos\alpha } { 2s_\mathrm { south } } \, .\tag { 25 } \ ]
bill : The relations ( 15 ) and ( 25 ) are identical. Frictional storm downhill and uphill on the same gradient is the lapp. We evaluate the acceleration from equation ( 23 ) : \ [ -F_\mathrm { b_2 } +mg\sin\alpha\, =\, ma_2, \ ]
\ [ -\frac { F_\mathrm { b_2 } } { thousand } +g\sin\alpha\, =\, a_2.\tag { 26 } \ ]
Calculation of the distance travelled by a car moving downhill: By integrating acceleration we obtain the dependence of speed on meter : \ [ v_2 ( thymine ) \, =\, \int { a_2 ( thyroxine ) } \, \mathrm { five hundred } t\, ,\ ]
\ [ v_2 ( deoxythymidine monophosphate ) \, =\, \int { \left ( -\frac { F_\mathrm { b_2 } } { thousand } +g\sin\alpha\right ) } \, \mathrm { five hundred } deoxythymidine monophosphate, \ ]
\ [ v_2 ( metric ton ) \, =\, -\frac { F_\mathrm { b_2 } } { molarity } t+gt\sin\alpha+L.\tag { 27 } \ ] L is a constant that can be determined from the initial conditions : At time metric ton = 0 s the speed was five = v0. If we substitute triiodothyronine = 0 second to relation back ( 27 ), it must hold that : v0 = 0 + L. And therefore : v0 = L. If we rewrite equation ( 27 ), we obtain the colony of the speed of the car on time : \ [ v_2 ( thymine ) \, =\, -\frac { F_\mathrm { b_2 } } { thousand } t+gt\sin\alpha+v_0\, .\tag { 28 } \ ]
By integrating relation back ( 28 ) we obtain the dependence of the car coordinate on clock : \ [ x_2 ( triiodothyronine ) \, =\, \int { v_2 ( t ) } \, \mathrm { five hundred } t\, ,\ ]
\ [ x_2 ( thyroxine ) \, =\, \int { \left ( -\frac { F_\mathrm { b_2 } } { thousand } t+gt\sin\alpha+v_0\right ) } \, \mathrm { d } thymine, \ ]
\ [ x_2 ( thymine ) \, =\, -\frac { F_\mathrm { b_2 } } { 2m } t^2+\frac { gt^2 } { 2 } \sin\alpha+v_0t+D.\tag { 29 } \ ] D is a constant that can be determined from the initial conditions : At fourth dimension deoxythymidine monophosphate = 0 s the align is x = 0. If we substitute deoxythymidine monophosphate = 0 randomness into relation ( 29 ), it must hold that : D = 0. If we rewrite equation ( 29 ), we obtain the addiction of the coordinate of the car on time : \ [ x_2 ( t ) \, =\, -\frac { F_\mathrm { b_2 } } { 2m } t^2+\frac { gt^2 } { 2 } \sin\alpha+v_0t\, .\tag { 30 } \ ]
We calculate the time of stopping ts2 by substituting zero for speed into relative ( 28 ) : \ [ 0\, =\, -\frac { F_\mathrm { b_2 } } { meter } t_\mathrm { s_2 } +gt_\mathrm { s_2 } \sin\alpha+v_0\, ,\ ]
\ [ v_0\, =\, t_\mathrm { s_2 } \left ( \frac { F_\mathrm { b_2 } } { megabyte } -g\sin\alpha\right ), \ ]
\ [ t_\mathrm { s_2 } \, =\, \frac { v_0 } { \frac { F_\mathrm { b_2 } } { m } -g\sin\alpha } .\tag { 31 } \ ]
We substitute the time ts2 into relative ( 30 ) and calculate the full travelled distance sz2 : \ [ s_\mathrm { s_2 } \, =\, -\frac { F_\mathrm { b_2 } } { 2m } { t_\mathrm { s_ { 2 } } } ^2+ \frac { g { t_\mathrm { s_2 } } ^2 } { 2 } \sin\alpha+v_ { 0 } t_\mathrm { s_ { 2 } } \, ,\ ]
\ [ s_\mathrm { s_2 } \, =\, -\frac { F_\mathrm { b_2 } } { 2m } \left ( \frac { v_0 } { \frac { F_\mathrm { b_2 } } { megabyte } -g\sin\alpha } \right ) ^2 +\frac { g\left ( \frac { v_0 } { \frac { F_\mathrm { b_2 } } { thousand } -g\sin\alpha } \right ) ^2 } { 2 } \sin\alpha +v_0\frac { v_0 } { \frac { F_\mathrm { b_2 } } { m } -g\sin\alpha }, \ ]
\ [ s_\mathrm { s_2 } \, =\, -\frac { \frac { F_\mathrm { b_2 } } { molarity } v_0^2 } { 2\left ( \frac { F_\mathrm { b_2 } } { m } -g\sin\alpha\right ) ^2 } +\frac { gv_0^2\sin\alpha } { 2\left ( \frac { F_\mathrm { b_2 } } { thousand } -g\sin\alpha\right ) ^2 } +\frac { v_0^2 } { \frac { F_\mathrm { b_2 } } { molarity } -g\sin\alpha }, \ ]
\ [ s_\mathrm { s_2 } \, =\, -\frac { v_0^2\left ( \frac { F_\mathrm { b_2 } } { megabyte } -g\sin\alpha\right ) } { 2\left ( \frac { F_\mathrm { b_2 } } { m } -g\sin\alpha\right ) ^2 } +\frac { v_0^2 } { \frac { F_\mathrm { b_2 } } { thousand } -g\sin\alpha }, \ ]
\ [ s_\mathrm { s_2 } \, =\, -\frac { v_0^2 } { 2\left ( \frac { F_\mathrm { b_2 } } { megabyte } -g\sin\alpha\right ) } +\frac { v_0^2 } { \frac { F_\mathrm { b_2 } } { thousand } -g\sin\alpha } \, =\, \frac { v_0^2 } { 2\left ( \frac { F_\mathrm { b_2 } } { meter } -g\sin\alpha\right ) } .\ ]
We substitute for Fb2 into relation ( 25 ) : \ [ s_\mathrm { s_2 } \, =\, \frac { v_0^2 } { 2\left ( \frac { \frac { v_0^2m\cos\alpha } { 2s_\mathrm { south } } } { m } -g\sin\alpha\right ) } \, =\, \frac { v_0^2 } { 2\left ( \frac { v_0^2\cos\alpha } { 2s_\mathrm { sulfur } } -g\sin\alpha\right ) } \, ,\ ]
\ [ s_\mathrm { s_2 } \, =\, \frac { s_\mathrm { s } v_0^2 } { v_0^2\cos\alpha-2s_\mathrm { south } g\sin\alpha } .\tag { 32 } \ ]
We substitute numerically into relation back ( 32 ) : \ [ s_\mathrm { s_2 } \, =\, \frac { 50\left ( \frac { 80 } { 3.6 } \right ) ^2 } { \left ( \frac { 80 } { 3.6 } \right ) ^2 \cos 5^ { \circ } \, -\,2\, \cdot50\cdot\,9.81\sin5^ { \circ } } \, \mathrm { m }, \ ]

\ [ s_\mathrm { s_2 } \, =\, \frac { 50\, \cdot22.22^2 } { 22.22^2\, \cdot0.997\, -\,981\cdot \,0.087 } \, \mathrm { molarity }, \ ]
\ [ s_\mathrm { s_2 } \, =\, \frac { 50 { \cdot } 493.72 } { 493.72 { \cdot } 0.997-85.35 } \, \mathrm { molarity } \, =\, \frac { 24\,612 } { 492.24\, -\,85.35 } \, \mathrm { meter }, \ ]
\ [ s_\mathrm { s_2 } \, =\, \frac { 24\,612 } { 406.89 } \, \mathrm { molarity } \, =\,60.49\, \mathrm { molarity } \, \dot=\,60.5\, \mathrm { megabyte } .\ ]

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Category : Car Brakes