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Answer:
The string of the kite is being let out at the rate of 19.2 ft/sec.
Step-by-step explanation:
See the diagram attached.
The vertical height (BK) of the kite is 120 ft.
The kite string is of length (AK) of 200 ft.
Therefore, the horizontal distance (AB) from Andrew to kite is = ft.
Now, applying the Pythagoras Theorem,
AB² + BK² = AK²
⇒ l² + h² = s² {Where, l is the horizontal length, h is the height and s is the string length}
Differentiating with respect to time t both sides
{Since, height of kite is constant}
⇒
⇒ ft per sec.
Therefore, when the wind is carrying the kite horizontally at the rate of 24 ft/sec, then the string of the kite is being let out at the rate of 19.2 ft/sec. (Answer)
Recall your d = rt, distance = rate * time
let's say we have two cars, A and B, B has a speed rate of "r", so A is the faster one and has a speed rate of " r + 10 "
they both are 300 miles apart after 3hrs, after 3hrs, car has been running for 3hrs, and car B has also been running for 3hrs, so their time is the same
we know their distances added up, is 300miles, so if car B covered say, "d" distance on those 3hrs, car A covered the slack, " 300 - d "
solve for "r"
let's say we have two cars, A and B, B has a speed rate of "r", so A is the faster one and has a speed rate of " r + 10 "
they both are 300 miles apart after 3hrs, after 3hrs, car has been running for 3hrs, and car B has also been running for 3hrs, so their time is the same
we know their distances added up, is 300miles, so if car B covered say, "d" distance on those 3hrs, car A covered the slack, " 300 - d "
solve for "r"