The distance, d, that an object falls is directly proportional to the square of the time, t, it has been in free fall. An object that has been in free fall for 7 seconds has fallen 784 feet. Determine the distance the object has fallen if it has been falling for 3 seconds.
We can use the principle of direct variation to set up a proportion:
d1/t1^2 = d2/t2^2
Given that d1 = 784 feet, t1 = 7 seconds, and t2 = 3 seconds, we can plug in these values into the equation:
784/7^2 = d2/3^2
Simplifying the equation:
784/49 = d2/9
Cross multiplying:
d2 = (784/49) * 9
d2 = 144 feet
Therefore, if the object has been falling for 3 seconds, it has fallen 144 feet.
To solve this problem, we can set up a proportion using the given information.
Let's assume that the distance the object has fallen after 3 seconds is represented by d1.
According to the problem, the distance is directly proportional to the square of the time:
d ∝ t^2
Using this, we can write the proportion as:
d/d1 = t^2/t1^2
where d is the distance fallen after 7 seconds (784 feet) and t is the time (7 seconds). t1 is the time after which we want to find the distance fallen (3 seconds).
Plugging in the known values, we have:
784/d1 = 7^2/3^2
Simplifying, we get:
784/d1 = 49/9
To solve for d1, we can cross-multiply:
49d1 = 784 * 9
Now, we can solve for d1 by dividing both sides by 49:
d1 = (784 * 9) / 49
Calculating this, we find that:
d1 = 144 feet
Therefore, if the object has been falling for 3 seconds, it will have fallen a distance of 144 feet.