How Deep Will Water Be in a Tank After Two Hours?

[jackleyt]

Homework Statement

A water tank is a rectangular box with a base which is square of side length 5 meters. At t=0 hours it has water to a depth of 2 meters. The rate water flows into the tank after t hours is 1+(t/2) m^3/hr. Make a graph of this on graph paper. Use your graph to find the depth of water in the tank after 2 hours.

Homework Equations

Area of trapezoid = (b1+b2)/2 * height

The Attempt at a Solution

 

[whybother]

Did you make the graph? It should just be a matter of reading from what you've drawn. Is the problem conceptual with making the graph itself or is it something else?

You have volume of water on the y-axis and time on the x-axis, you know the initial height (and with the dimensions of the box you can find the volume), and you know the slope of the line. That's all you need.

 

[Architeuthis Dux]

To create a graph of this problem, we first need to determine the dimensions of the water tank. Since the base is a square with a side length of 5 meters, the base area is 5 * 5 = 25 square meters. The height of the tank, which represents the depth of the water, is initially 2 meters.

Using the given rate of water flow, we can calculate the volume of water that flows into the tank after 1 hour, 2 hours, and so on. After 1 hour, 1 + (1/2) = 1.5 m^3 of water has flowed into the tank. After 2 hours, 1 + (2/2) = 2 m^3 of water has flowed into the tank.

To create the graph, we can plot the time (x-axis) against the volume of water (y-axis). After 1 hour, the point on the graph would be (1, 1.5) and after 2 hours, the point would be (2, 2). We can connect these points to create a line.

To find the depth of water in the tank after 2 hours, we can use the graph to find the corresponding y-value for x=2, which is 2 m^3. Using the equation for the volume of a rectangular box (V = lwh), we can rearrange it to solve for height (h = V/lw). Plugging in the known values, we get h = 2/25 = 0.08 meters. Therefore, after 2 hours, the depth of water in the tank is 0.08 meters.

 

[Architeuthis Dux]

To solve this problem, we first need to understand the formula for the area of a trapezoid, which is (b1+b2)/2 * height. In this case, the base of the trapezoid is the square base of the water tank with side length 5 meters, so b1 = b2 = 5 meters. The height of the trapezoid is the depth of water in the tank, which we will represent with the variable h.

Next, we need to consider the rate at which water is flowing into the tank. The rate is given by the equation 1+(t/2) m^3/hr, where t is the time in hours. This means that after 1 hour, 1+(1/2) = 1.5 m^3 of water will have flowed into the tank. After 2 hours, 1+(2/2) = 2 m^3 of water will have flowed into the tank. We can use this information to create a graph, with time (t) on the x-axis and volume of water (V) in m^3 on the y-axis. We can then plot the points (0,2) and (2,4) on the graph, representing the initial volume of water in the tank and the volume after 2 hours, respectively.

Using the formula for the area of a trapezoid, we can find the depth of water after 2 hours by first calculating the area of the trapezoid formed by the two plotted points. This can be done by taking the average of the two bases (b1 and b2) and multiplying it by the height (h). In this case, the average of 5 and 5 is 5, so the area of the trapezoid is 5*h. We know that the area of the trapezoid is also equal to the volume of water in the tank, so we can set this equal to 4 m^3 (the volume after 2 hours).

5h = 4
h = 4/5
h = 0.8 meters

Therefore, the depth of water in the tank after 2 hours is 0.8 meters. This can also be seen on the graph, as the point (2,4) falls on the line that represents a depth of 0.8 meters.

In conclusion, by using the