Question

two jets leave an air base at the same time and travel in opposite directions. one jet travels 71 mih slower than the other. if the two jets are 5764 miles apart after 4 hours, what is the rate of each jet?

Answer 1

Answer:

Speed of Faster jet is 756 miles/hr and speed of slower jet is 685 miles/hr.

Step-by-step explanation:

Let the speed of faster jet be represent as 's'

Now Given:

one jet travels 71 mih slower than the other.

Hence Speed of slower jet will be = $s-71$

Distance = 5764 miles

Time = 4 hrs

Now we know that;

Distance is equal to product of speed and time.

Framing in equation for we get;

Distance = (Speed of Faster Jet + Speed of Slower jet) × Time.

Substituting the given values we get;

$5764=(s+s-71)\times 4\\\\5764= (2s-71)\times 4\\\\\frac{5764}{4} = 2s-71\\\\1441=2s-71\\\\2s=1441+71\\\\2s =1512\\\\s =\frac{1512}{2} = 756\ mi/h$

Speed of faster jet = 756 miles/hr

Speed of slower jet = $s-71 =756-71 = 685\ mi/hr$

Hence Speed of Faster jet is 756 miles/hr and speed of slower jet is 685 miles/hr.

Now we will check the answer;

Distance traveled by faster jet = speed × time = 756 × 4 = 3024 miles.

Distance traveled by Slower jet = speed × time = 685 × 4 = 2740 miles

Hence Total Distance = 3024 + 2740 = 5764 miles.