91Ó°ÊÓ

Consider that the entire motion of the car is in a single direction. Also, the distance of sliding is related to the variation in kinetic energy of the car.

Given data:

The mass of car A is\({m_{\rm{A}}} = 1500\;{\rm{kg}}\).

The mass of car B is\({m_{\rm{B}}} = 1100\;{\rm{kg}}\).

The distance at which brakes are applied is\(d = 15\;{\rm{m}}\).

The slide distance of A is\({s_{\rm{A}}} = 18\;{\rm{m}}\).

The slide distance of B is\({s_{\rm{B}}} = 30\;{\rm{m}}\).

The coefficient of friction is\(\mu = 0.60\).

The speed limit is \({v_{\rm{L}}} = 90\;{\rm{km/h}}\).

The relation to calculate work done can be written as follows:

\(\begin{array}{c}W = \Delta E\\ - {F_{\rm{f}}} \times s = \frac{1}{2}m\left( {{{v'}^2} - {v^2}} \right)\\ - \mu mg \times s = \frac{1}{2}m\left( {{{v'}^2} - {v^2}} \right)\\ - \mu g \times s = \frac{1}{2}\left( {{{v'}^2} - {v^2}} \right)\end{array}\) … (i)

Here,\({F_{\rm{f}}}\)is the frictional force; s is the distance moved by car; mis the mass of the car;\(\Delta E\)is the change in kinetic energy; g is the gravitational acceleration; vand ±¹â€™ are the initial and final speed after a collision, where the final speed is zero.

Plugging the values in the above equation to find the speed of car A,

\(\begin{array}{c} - \mu g \times s = \frac{1}{2}\left( {{{\left( 0 \right)}^2} - {{v'}_{\rm{A}}}^2} \right)\\ - \left( {0.60} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {18\;{\rm{m}}} \right) = \frac{1}{2}\left( {{{\left( 0 \right)}^2} - {{\left( {{{v'}_{\rm{A}}}} \right)}^2}} \right)\\{{v'}_{\rm{A}}} = 14.5\;{\rm{m/s}}\end{array}\)

Plugging the values in the above equation to find the speed of car B,

\(\begin{array}{c} - \mu g \times s = \frac{1}{2}\left( {{{\left( 0 \right)}^2} - {{v'}_{\rm{B}}}^2} \right)\\ - \left( {0.60} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {30\;{\rm{m}}} \right) = \frac{1}{2}\left( {{{\left( 0 \right)}^2} - {{\left( {{{v'}_{\rm{B}}}} \right)}^2}} \right)\\{{v'}_{\rm{B}}} = 18.7\;{\rm{m/s}}\end{array}\)

The relation from conservation of momentum can be written as follows:

\(\begin{array}{c}{P_{\rm{i}}} = {P_{\rm{f}}}\\{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}{{v'}_{\rm{B}}}\\{m_{\rm{A}}}{v_{\rm{A}}} + 0 = {m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}{{v'}_{\rm{B}}}\\{v_{\rm{A}}} = \frac{{{m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}{{v'}_{\rm{B}}}}}{{{m_{\rm{A}}}}}\end{array}\)

Here, \({v_{\rm{B}}}\) is the initial velocity of car B, whose value is zero.

Plugging the values in the above equation,

\(\begin{array}{l}{v_{\rm{A}}} = \left[ {\frac{{\left( {1500\;{\rm{kg}}} \right)\left( {14.5\;{\rm{m/s}}} \right) + \left( {1100\;{\rm{kg}}} \right)\left( {18.7\;{\rm{m/s}}} \right)}}{{1500\;{\rm{kg}}}}} \right]\\{v_{\rm{A}}} = 28.3\;{\rm{m/s}}\end{array}\).

The equation (i) can be written in the following form:

\(\begin{array}{c} - \mu gd = \frac{1}{2}\left( {{v_{\rm{A}}}^2 - {v^2}} \right)\\v = \sqrt {{v_{\rm{A}}}^2 + 2\mu gd} \end{array}\)

Here, v is the speed of car A exceeding the speed limit.

Plugging the values in the above equation,

\(\begin{array}{l}v = \sqrt {{{\left( {28.3\;{\rm{m/s}}} \right)}^2} + 2\left( {0.60} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {15\;{\rm{m}}} \right)} \\v = \left( {31.2\;{\rm{m/s}} \times \frac{{1\;{\rm{mi/h}}}}{{0.447\;{\rm{m/s}}}}} \right)\\v = 69.7\;{\rm{mi/h}}\end{array}\)

Thus, \(v = 69.7\;{\rm{mi/h}}\) is the speed of the car exceeding the speed limit.