Travelling at 60 km/h , a car overtakes another car travelling at...

The speeds of the two cars in $\text{m/s}$ are:
$v_A = 60 \times \dfrac{5}{18} = \dfrac{50}{3} \text{ m/s}$
$v_B = 42 \times \dfrac{5}{18} = \dfrac{35}{3} \text{ m/s}$

The relative speed of car A with respect to car B is:
$v_{rel} = v_A - v_B = \dfrac{50}{3} - \dfrac{35}{3} = 5 \text{ m/s}$

For car A to completely overtake car B, the front of car A must travel from the rear of car B to a point where the rear of car A is just ahead of the front of car B. The total relative distance to be covered is the sum of the lengths of both cars:
$d_{rel} = L_A + L_B = 5.0 + 5.0 = 10.0 \text{ m}$

The time taken for the overtake is:
$t = \dfrac{d_{rel}}{v_{rel}} = \dfrac{10.0}{5} = 2 \text{ s}$

During this time, the actual distance travelled by car A is:
$d_A = v_A \times t = \dfrac{50}{3} \times 2 = \dfrac{100}{3} \approx 33.33 \text{ m}$

The total road distance used for the overtake is the total length of the road occupied during the maneuver, which is measured from the initial position of the rear of car A to the final position of the front of car A. This is equivalent to the distance travelled by car A plus its own length:
$\text{Total road distance} = d_A + L_A = 33.33 + 5.0 = 38.33 \text{ m}$

Rounding to the nearest integer, the total road distance used is $38 \text{ m}$.

Answer: $2 \text{ s}$, $38 \text{ m}$