the equation of line t is 6x+5y=3, and the equation of line q is 5x-6y=0. Which statement about the two lines is true?
1 answer:
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Answer:
-3/2+√29/2 or -3/2-√29/2
Step-by-step explanation:
When the graph intercepts with x-axis, the value of y is zero.
So substitute y with zero.
Answer:
We assume that the orange is dropped at t = 0s.
Once the orange is on the air, the only force acting on it is the gravitational force, then the acceleration of the orange is the gravitational acceleration.
A(t) = -32.17 ft/s^2
Where the negative sign is because this acceleration points downwards.
For the velocity equation, we need to integrate over time, we will get:
V(t) = (-32.17 ft/s^2)*t + V0
Where V0 is the initial vertical velocity of the orange, because the orange is accidentally dropped, this initial velocity is equal to zero.
V(t) = (-32.17 ft/s^2)*t
For the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + P0
Where P0 is the initial height of the orange, we know that it is 40ft, then the position equation is:
P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
Now that we know the equation, we can graph it. (you can see the graph below)
Now we also want to find at what time does the orange hit the water.
This happens when:
P(t) = 0 ft = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
We just need to solve that equation for t.
0 ft = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
(1/2)*(32.17 ft/s^2)*t^2 = 40 ft
t^2 = (40ft)/( (1/2)*(32.17 ft/s^2))
t = √( (40ft)/( (1/2)*(32.17 ft/s^2)) ) = 1.58 s
The orange hits the water 1.58 seconds after it is dropped.
