SOLUTION: I don't have any idea how to do these word problems. I've been trying but I just don't get it. Can you please help me? I can never understand word problems.
1) The distance an o
Question 161461: I don't have any idea how to do these word problems. I've been trying but I just don't get it. Can you please help me? I can never understand word problems.
1) The distance an object falls is directly proportional to the square of the time it has been falling. After 6 seconds it has fallen 1296 feet. How longwill it take to fall 2304 feet?
2) x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?
If you can help I'd really appreciate it. Thank you so much! I could not do this without you. Answer by KnightOwlTutor(293) (Show Source):
You can put this solution on YOUR website! 1. distance= aXtime squared. we know that distance is proportinal but we don't know what a is. That is why they gave you the time and the distance in the first part of the equation. This will solve for the unknown a so that you can proceed to the second part of the problem.
1296=a(6)^2 1296=36a
divide both sides by 36
a=12
We have the distance and a so we can solve for t
2304=12(t)^2
Divide both sides by 12
192=t^2
Take the square root
t=13.85 seconds
Let's check by plugging into the equation to see if we are right
12(13.9)^2=2,318.52
It is a close approximation for the problem.
2) x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?
x=s^2/t We know that when something varies directly. it is like this a=b
When something varies indirectly it looks like this a=1/b
when s is doubled
x=(2s)^2/t x=4s^2/t x is 4x as great.
When both s and t are doubled
x=(2S)^2/2t= x=4s^2/2t x=2s^2/t the value x is doubled
