(a)
The total energy used by car is,
\({E_{out}} = \frac{1}{2}m{v^2} + mgh + fd\)
Here, \(m\) is the mass of the car \(\left( {m = 900{\rm{ kg}}} \right)\), \(v\) is the velocity of car \(\left( {v = 30.0{\rm{ m}}/{\rm{s}}} \right)\), \(h\) is the gain in altitude \(\left( {h = 3.00{\rm{ km}}} \right)\), \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), \(f\) is the average force of friction \(\left( {f = 700{\rm{ N}}} \right)\), and \(d\) is the distance traveled by the car \(\left( {d = 100{\rm{ km}}} \right)\).
Putting all known values,
\(\begin{array}{c}{E_{out}} = \left[ \begin{array}{l}\frac{1}{2} \times \left( {900{\rm{ kg}}} \right) \times {\left( {30{\rm{ m}}/{\rm{s}}} \right)^2} + \left( {900{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3{\rm{ km}}} \right)\\ + \left( {700{\rm{ N}}} \right) \times \left( {100{\rm{ km}}} \right)\end{array} \right]\\ = \left[ \begin{array}{l}\frac{1}{2} \times \left( {900{\rm{ kg}}} \right) \times {\left( {30{\rm{ m}}/{\rm{s}}} \right)^2} + \left( {900{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3{\rm{ km}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\\ + \left( {700{\rm{ N}}} \right) \times \left( {100{\rm{ km}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\end{array} \right]\\ = 9.6865 \times {10^7}{\rm{ J}}\end{array}\)
The energy input of \(1{\rm{ gal}}\) of gasoline is,
\({E_{in}} = 1.2 \times {10^8}{\rm{ J}}\)
The efficiency of the car is,
\(\eta \% = \frac{{{E_{out}}}}{{{E_{in}}}} \times 100\% \)
Putting all known values,
\(\begin{array}{c}\eta = \frac{{9.6865 \times {{10}^7}{\rm{ J}}}}{{1.2 \times {{10}^8}{\rm{ J}}}} \times 100\% \\ = 80.72\% \end{array}\)
Therefore, the required efficiency of the car is \(80.72\).
