Laura lives 15 miles east of Kevin’s place. Kevin lives 8 miles south of Michelle’s place. How far does Michelle live from Laura
1 answer:
Answer:
17 miles.
Step-by-step explanation:
Let's define:
North as the positive y-axis
East as the positive x-axis.
We know that Laura lives 15 miles east of Kevin's place.
Kevin lives 8 miles south of Michelle's place.
So, if we define the origin, (0, 0) as Laura's place.
From:
"that Laura lives 15 miles east of Kevin's place."
We have that the location of Kevin's house is 15 miles west from Laura's place, then Kevin's house is at:
(0, 0) + (-15mi, 0) = (-15mi, 0)
From Kevin lives 8 miles south of Michelle's place, we know that Michelle's live 8 miles north of Kevin's place.
Then the location of Michele's house is the location of Kevin's plus (0, 8mi).
Michelle's house is located at:
(-15mi, 0) + (0, 8mi) =(-15mi, 8mi)
Now we want to find the distance between Michelle's house and Laura's house.
Michelle's house is at (-15mi, 8mi)
Laura's house is at (0mi, 0mi)
Remember that the distance between two points (a, b) and (c, d) is given by:
Then the distance between (-15mi, 8mi) and (0mi, 0mi) is:
The correct option is the first one, 17 miles.
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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.
