2 Integration problems

[hips_dont_ly]

Hello! So here are two problems that I believe are integration problems, but the methods at which I am doing them seem to keep running me into dead ends. And hopefully you guys won't think I'm just listing homework problems; these are the last two that my roommates and friends can't solve either.

1) A car accelerates from 0mph to 40 mph. Its velocity at time t is v(t)=5t mph where t is in seconds. Use an integral to find the total distance in miles it travels during its acceleration. (Hint: work with seconds not hours.)
---So for this problem, it blatantly told me to use an integral to find the problem, which I found to be 5t^2/2 or 2.5t^2
---After that, I really didn't know exactly where else to begin. I thought that v(t)=5t simply means that the velocity at any time is simply 5 times the time.
---I attempted to convert whatever I was given into seconds. 40 miles per hour became .6666667 miles per minute, becoming .0111111 miles per second.
---Using the integration symbol ?, I set my limits as 0 and 40, and found that the answer was about 3.086 e-4
---Incorrect answer. I then realized that I did not know how to find how fast it was accelerating from 0 to 40. Any help?

2) A water tank has a square base with each side of length 5 meters. Water enters through a hose at a constant rate of 20 liters per minute. At the same time a valve in the bottom is opening, so that after t minutes, water leaves at a rate of t liters per minute. If the tank starts out filled to a depth of 4 meters, after how many minutes will the tank be empty?
---So I began thinking that my goal was to basically find a formula for say h(t), the depth of water after t minutes.
---I found the area of my base, 5^2, to be 25. Using that, my volume was 25h(t)
---Because water is leaving the tank, I set an equation as 25h'(t)= -t so h'(t) was -t/25
---Integrating that gave me -t^2/27
Because at t(0)=4, h(t)=4 - t^2/27. To sum up I eventually solved for t for time and that was incorrect as well.

I'd appreciate any help guys! Thank you.

 

[wjm11]

1) A car accelerates from 0mph to 40 mph. Its velocity at time t is v(t)=5t mph where t is in seconds. Use an integral to find the total distance in miles it travels during its acceleration. (Hint: work with seconds not hours.)
---So for this problem, it blatantly told me to use an integral to find the problem, which I found to be 5t^2/2 or 2.5t^2
---After that, I really didn't know exactly where else to begin. I thought that v(t)=5t simply means that the velocity at any time is simply 5 times the time.
---I attempted to convert whatever I was given into seconds. 40 miles per hour became .6666667 miles per minute, becoming .0111111 miles per second.
---Using the integration symbol ?, I set my limits as 0 and 40, and found that the answer was about 3.086 e-4
---Incorrect answer. I then realized that I did not know how to find how fast it was accelerating from 0 to 40. Any help?

Lots of ways to do this. Here is one. Start by drawing a velocity graph.

Your velocity is a linear function that plots through the origin with a slope of 5. The x-axis is “t” (with units in seconds). Your y-axis is velocity (with units in mph) Since we’re interested in the 0 to 40 mph (y-axis), that means we’re looking at 0<t<8 on the x-axis. The area under the curve represents the displacement. The area is a triangular region with height “40 mph” and base “8 s”. Change the “8 s” into “8/3600 hr”.

Area (displacement) = (.5(40 mi/hr)(8hr/3600) = .044… miles

PS You can use the integration approach, too, but you need to integrate from 0 to 8 (x-axis) and convert the mph into mps (which will change your v(t) function).

 

[wjm11]

A water tank has a square base with each side of length 5 meters. Water enters through a hose at a constant rate of 20 liters per minute. At the same time a valve in the bottom is opening, so that after t minutes, water leaves at a rate of t liters per minute. If the tank starts out filled to a depth of 4 meters, after how many minutes will the tank be empty?
---So I began thinking that my goal was to basically find a formula for say h(t), the depth of water after t minutes.
---I found the area of my base, 5^2, to be 25. Using that, my volume was 25h(t)
---Because water is leaving the tank, I set an equation as 25h'(t)= -t so h'(t) was -t/25
---Integrating that gave me -t^2/27
Because at t(0)=4, h(t)=4 - t^2/27. To sum up I eventually solved for t for time and that was incorrect as well.

Once again, you must pay attention to units. 1L/min = 10^-3 cubic meters per minute (please verify this for yourself). We have both an inflow and an outflow of water in this problem, so

dV/dt = (20)(10^-3) – (10^-3)t (m^3/min)

This is the rate of change of volume with respect to time. Make sense? Now integrate to get the volume versus time function:

V(t) = (20)(10^-3)t – (.5)(10^-3)t^2 (m^3) + C

C = V(0) = 4(5)(5) = 100 m^3

V(t) = (20)(10^-3)t – (.5)(10^-3)t^2 (m^3) + 100 = 0

Solve.

 

[hips_dont_ly]

Thank you! For the second, problem,
"This is the rate of change of volume with respect to time. Make sense? Now integrate to get the volume versus time function:

V(t) = (20)(10^-3)t – (.5)(10^-3)t^2 (m^3) + C"

I was unclear with setting up the rate of change to respect to time. I'm assuming that it implies that the rate of change is just the derivative of the water level?

Once again, thank you for helping me with my problems

 

[wjm11]

I'm assuming that it implies that the rate of change is just the derivative of the water level?

No. Please reread what I wrote: "This is the rate of change of VOLUME with respect to time." I did not say anything about height.

"We have both an inflow [(20)(10^-3)] and an outflow [– (10^-3)t] of water:

dV/dt = (20)(10^-3) – (10^-3)t (m^3/min)"

I integrated dV/dt. I then solved for "C."