Airplane at 19,200 feet descending at a rate of 40 feet per second. Another airplane takes off and ascends at a rate of 60 feet per second. After how many seconds will the airplanes be at the same height? What is this height?
19,200 - 40s = 0 + 60s
Solve for s, then 19,200-40s will be altitude.
192
To find the number of seconds it takes for the airplanes to be at the same height, we need to calculate the height of each airplane at any given time and compare them.
Let's start by finding the height of the descending airplane at time t (in seconds). We know that the initial height is 19,200 feet, and it descends at a rate of 40 feet per second. Therefore, the height of the descending airplane at time t is given by:
Height of descending airplane = 19,200 - (40 * t) (Equation 1)
Next, let's find the height of the ascending airplane at time t. We know that the initial height is 0 feet (since it just took off), and it ascends at a rate of 60 feet per second. Therefore, the height of the ascending airplane at time t is given by:
Height of ascending airplane = 0 + (60 * t) (Equation 2)
To find when the airplanes will be at the same height, we need to set Equation 1 equal to Equation 2 and solve for t:
19,200 - (40 * t) = 60 * t
Now, let's solve the equation:
19,200 = 60 * t + 40 * t (combining like terms)
19,200 = 100 * t (combining like terms)
t = 192
Thus, it will take 192 seconds for the airplanes to be at the same height.
To find the height at that time, we substitute t = 192 into either Equation 1 or Equation 2:
Height of descending airplane = 19,200 - (40 * 192)
= 19,200 - 7,680
= 11,520 feet
Therefore, after 192 seconds, both airplanes will be at the height of 11,520 feet.