the distance covered by a falling ball varies as the square of the time taken. If a ball falls through 1800m in 15seconds,calculate the distance when the time taken is 15seconds and the time taken for is distance of 648m.solve using variation
Plss can u plss solve it to my own understanding
Am still in s.s 1
I don't understand plss
Thanks
Pls solve and explain to my own understanding i'm in sss1
To solve this problem using variation, we need to determine the relationship between the distance covered by the falling ball and the time taken.
Given that the distance covered varies as the square of the time taken, we can express this relationship mathematically as:
distance = k * time^2
Where "k" is a constant.
To find the value of "k", we can use the given information that the ball falls through 1800m in 15 seconds:
1800 = k * 15^2
Simplifying this equation, we have:
1800 = 225k
Dividing both sides of the equation by 225, we find:
k = 1800/225
k = 8
Now that we have the value of "k", we can determine the distance covered when the time taken is 15 seconds and the time taken for a distance of 648m.
For the first scenario:
distance = k * time^2
distance = 8 * 15^2
distance = 8 * 225
distance = 1800m
For the second scenario:
distance = k * time^2
648 = 8 * time^2
Dividing both sides of the equation by 8, we find:
81 = time^2
Taking the square root of both sides, we have:
time = √81
time = 9 seconds
Therefore, when the time taken is 9 seconds, the distance covered by the falling ball is 648m.
d = kt^2
So, d/t^2 is constant. You want d such that
648/t^2 = 1800/15^2
distance for 15 seconds is clearly 1800m, as they told you!