Question

An aircraft carrier left Port 35 traveling north five hours before a container ship. The container ship traveled in the opposite direction going 8 km/h faster than the aircraft carrier for seven hours after which time the ships were 303 km apart. What was the aircraft carrier's speed?

Answer 1

The aircraft carrier's speed was 13 km/h

Explanation:

The parameters given in the question are;

Time of departure of the aircraft carrier = 5 hours before the container ship

Speed of the container ship = 8 km/h faster than the aircraft carrier

Distance between the two ships after 7 hours = 303 km

Let the speed of the aircraft carrier be Z

∴ The speed of the container ship = Z + 8

Since the aircraft carrier left port 35 travelling North 5 hours earlier we have;

Total time of travel of the aircraft carrier at the time of measurement of the distance between the two ships = 5 + 7 = 12 hours

Therefore, since both ships are moving in opposite direction on the same line path, we have;

303 = 12×Z + 7 × (Z + 8) = 12·Z + 7·Z + 56

19·Z = 303 - 56 = 247

∴ Z = 247/19 = 13 km/h

Hence, the aircraft carrier's speed was 13 km/h.

Answer 2

21.75 km/h

Explanation:

Let's call the aircraft carrier speed 'x', so the speed of the container ship is 'x+8'

The distance travelled can be found with the product of the speed and the time travelled: d = v * t

The aircraft carrier travelled five hours, so the distance travelled is:

d1 = x * 5 = 5x

Then, the container ship travelled seven hours, so the distance travelled is:

d2 = (x+8)*7 = 7x + 42

As they travelled in opposite directions, the distance between then is the sum of these distances, and it is equal 303 km, so we have that:

d1 + d2 = 303

5x + 7x + 42 = 303

12x = 261

x = 21.75 km/h