Question

A radar locates an enemy bomber 400 km away and flying directly toward the capital city at a speed of 600 km per hour. At the same moment, a fighter plane leaves the capital city at a speed of 750 km per hour. When and where will the fighter plane intersect the enemy bomber?

Answer 1

Question #35699

A radar locates an enemy bomber 400 km away and flying directly toward the capital city at a speed of 600 km per hour. At the same moment, a fighter plane leaves the capital city at a speed of 750 km per hour. When and where will the fighter plane intersect the enemy bomber?

Solution:

Let

$$S _ {0} = 4 0 0 \mathrm {k m}$$ $$v _ {1} = 6 0 0 \mathrm{km/h}$$ $$v _ {2} = 7 5 0 \mathrm{km/h}$$ $$S = ?, t _ {x} = ?$$

The distance between the bomber and the capital city defined as

$$S = S _ {0} - v _ {1} t \text{ were } t \text{ is the time}$$

The distance between the fighter plan and the capital city defined as

$$S = v _ {2} t \text{ were } t \text{ is the time}$$

When the fighter plan intersect the bomber (through time $t_x$) their distances is equal

$$S _ {0} - v _ {1} t _ {x} = v _ {2} t _ {x}$$ $$t _ {x} = \frac {S _ {0}}{v _ {1} + v _ {2}}$$

The distance from the city at this time is

$$S = t _ {x} v _ {2}$$ $$t _ {x} = \frac {4 0 0}{6 0 0 + 7 5 0} = 0.3 \mathrm{h}$$ $$S = 0.3 * 7 5 0 = 2 2 5 \mathrm{km}$$

Answer: through 0.3 h (or 18 min) at distance 225 km from city.