The sports car has a mass of 1.5 Mg Mg and a center of mass at G. Determine the shortest time it takes for it to reach a speed of 80 km/h km/h, starting from rest, if the engine only drives the rear wheels, whereas the front wheels are free rolling. The coefficient of friction between the wheels and road is μ=0.2μ=0.2. Neglect the mass of the wheels for the calculation.

Respuesta :

Answer:

[tex]\Delta t = 17.041\,s[/tex]

Explanation:

The figure of the problem is included below as attachment. The equations of equilibrium are presented below:

[tex]\Sigma F_{x} = -\mu\cdot N_{1} = m\cdot a[/tex]

[tex]\Sigma F_{y} = N_{1} + N_{2} - m\cdot g = 0[/tex]

[tex]\Sigma M_{G} = N_{1}\cdot x_{1} - N_{2}\cdot x_{2} - \mu\cdot N_{1}\cdot y_{1} = 0[/tex]

The system of equations are:

[tex]-0.2\cdot N_{1} = (1500\,kg)\cdot a[/tex]

[tex]N_{1} + N_{2} - (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) = 0[/tex]

[tex]N_{1} \cdot [(0.75\,m)-0.2\cdot (0.35\,m)] - N_{2}\cdot (1.25\,m) = 0[/tex]

The solution of the system is:

[tex]N_{1} = 9780.918\,N[/tex], [tex]N_{2} = 4929.582\,N[/tex] and [tex]a = - 1.304\,\frac{m}{s^{2}}[/tex]

The shortest time to reach a speed of 80 km/h is:

[tex]\Delta t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}}{1.304\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta t = 17.041\,s[/tex]

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