Maximum Velocity for different accelerations

[Noah Drakes]

Homework Statement

Force of jet= A(r(t))4/3
A: constant determined by the fighter model in the class being considered and the drag force on the plane
r(t): the rate of fuel consumption as a function of time

Consider 3 possible situations for r(t):

1. when the rate is constant for the duration of the acceleration period
2. when the rate is steadily decreasing for the duration of the acceleration period
3. when the rate decreases at a decreasing rate as the plane accelerates

For each of these you should assume that the rate is initially 500 ffu (fighter fuel units)/min,
and for each it is reasonable to suppose an acceleration period of 10 minutes.

Determine-in terms of the characteristics of the fighter (m and A)- the maximum velocity obtained by the jet

Homework Equations

F=ma

The Attempt at a Solution

I set F=ma equal to Fjet = A(r(t)^4/3. So ma=A(r(t)^4/3 .

Now i solved for "a" to get an equation for acceleration: a=(A(r(t)^4/3)/(m)

Then i differentiated the equation to get velocity : A/m * 3/7 (r(t) ^ (7/3)

confused on how to approach the problem from here. I have a test on wednesday with a problem similar to this problem. It would be immensly helpful if someone solved the problem explaining each step and why.

I am new to this community by the way.

 

[berkeman]

It would be immensly helpful if someone solved the problem explaining each step and why.

I am new to this community by the way.

It is apparent that you are new. We absolutely do not provide solutions to homework-type questions. We can provide hints and ask probing questions and find mistakes, but doing your work for you is verboten.

Can you say what differences there are between the 3 questions in your problem statement? How do the differences help you to approach the overall problem?

 

[Noah Drakes]

Ok i understand.

The differences between each statement is the state of (r(t)). It is either constant, increasing, or decreasing.
So that implies i would have to manipulate the value of (r(t)). When (r(t)) is constant its value remains at 500. When the rate is decreasing steadily, however, i would subtract a small value from 500 ffu for a few increments. So i would plot the values when (r(t)) = 490, 4980, 470, 460 and so on ? Am I on the right track.

 

[Noah Drakes]

Ok i understand.

The differences between each statement is the state of (r(t)). It is either constant, increasing, or decreasing.
So that implies i would have to manipulate the value of (r(t)). When (r(t)) is constant its value remains at 500. When the rate is decreasing steadily, however, i would subtract a small value from 500 ffu for a few increments. So i would plot the values when (r(t)) = 490, 4980, 470, 460 and so on ? Am I on the right track.

But this doesn't seem right. The jets' fuel consumption would never stay constant during an acceleration period. Am i misinterpreting the meaning of (r(t))?

 

[haruspex]

Then i differentiated the equation to get velocity

Differentiating acceleration will give jerk, not velocity. Anyway, it looks like you tried to integrate, but the expression you wrote for the integral would only be correct if r(t)=t.
You need to pick one of the three cases, rewrite the expression for a to match that case, and then integrate.

The jets' fuel consumption would never stay constant during an acceleration period

Why not?

 

[Noah Drakes]

Differentiating acceleration will give jerk, not velocity. Anyway, it looks like you tried to integrate, but the expression you wrote for the integral would only be correct if r(t)=t.
You need to pick one of the three cases, rewrite the expression for a to match that case, and then integrate.

Why not?

Haha whoops.

 

[Noah Drakes]

Differentiating acceleration will give jerk, not velocity. Anyway, it looks like you tried to integrate, but the expression you wrote for the integral would only be correct if r(t)=t.
You need to pick one of the three cases, rewrite the expression for a to match that case, and then integrate.

Why not?

Ok so if r(t) stays constant, and the initial value for r(t) is 500 ffu, then i should integrate the equation (A/m) * 500^4/3. but would i integrate with respect to A or m ?

 

[haruspex]

Ok so if r(t) stays constant, and the initial value for r(t) is 500 ffu, then i should integrate the equation (A/m) * 500^4/3. but would i integrate with respect to A or m ?

You are obtaining a velocity from an acceleration. What variable connects those?

 

[Noah Drakes]

Ok so if r(t) stays constant, and the initial value for r(t) is 500 ffu, then i should integrate the equation (A/m) * 500^4/3. but would i integrate with respect to A or m ?

and why am i integrating, to get the maximum velocity ?

 

[haruspex]

and why am i integrating, to get the maximum velocity ?

If an object moving at velocity u at time t₀ has acceleration a=a(t), what is its velocity at time t₁?
If a(t)>0 all that time, when does the maximum velocity occur?

 

[Noah Drakes]

To find the velocity you would take the integral of a (V = ∫ a(t) ) then plug t₁ into the equation for V.

the max velocity occurs when a = 0

 

[haruspex]

To find the velocity you would take the integral of a (V = ∫ a(t) ) then plug t1 into the equation for V.

Yes, except that you have not factored the variables u and t₀ that I specified into that.

the max velocity occurs when a = 0

I specified a>0 throughout the interval.

 

[Noah Drakes]

Yes, except that you have not factored the variables u and t₀ that I specified into that.

I specified a>0 throughout the interval.

V = (u) + (a * t) ?

 

[haruspex]

V = (u) + (a * t) ?

That is only for constant acceleration. Combine the two attempts, and don't forget to specify both bounds on the integral.

 

[Noah Drakes]

i feel stupid. i know this isn't a hard problem. i feel like I am wasting your time

 

[Noah Drakes]

∫_t0 ^t1(v-t)/u

 

[Noah Drakes]

You are obtaining a velocity from an acceleration. What variable connects those?

time

 

[haruspex]

∫_t0 ^t1(v-t)/u

The bounds are right, but the rest makes little sense. You cannot subtract a time from a velocity, and dividing by the initial velocity??
Go back to the integral you had in post #11. Fill in the variable of integration and the bounds. That is the generalised form of a*(t₁-t₀) when a is varying. It remains to use u the same way that you did in post#13.

 

[Noah Drakes]

∫_t0^t1 = u + a * (t₁ - t₀) da

 

[haruspex]

∫_t0^t1 = u + a * (t₁ - t₀) da

You do not seem to understand the fundamentals of integration. The expression on the right is only for constant acceleration, and anyway does not correspond to the expression on the left because the one on the left does not include u.

If the acceleration is constant from t₀ to t₁ then the gain in velocity is a(t₁-t₀). If you draw an acceleration-time graph it will be the area of the rectangle formed by the graph, the t axis and the two vertical lines t=t₀ and t=t₁, i.e. it will be a(t₁-t₀).

If the acceleration is not constant then you still want the area bounded by the graph of a and those other three lines, but it is no longer a simple rectangle. The principle of integration is that we slice up the shape into very small time intervals, i.e. thin vertical strips. Each strip can be approximated as a rectangle. Then we add all these rectangles together. In notation, we write this as ##\int_{t_0}^{t_1}a.dt##. As in the constant case, this gives us the increase in velocity from time t₀ to t₁, so to get the velocity at t₁ we need to add the velocity at t₀.

With that explanation, can you now write the correct formula for the variable acceleration case?

 

[Noah Drakes]

Ok, I think i Got it
So the equation for velocity when acceleration varies from t₀ to t₁ is V=∫_t0 ^t1 (A/m) (r(t)) ^3/4

 

[Noah Drakes]

where t₁ = 10 and t₀ is 0

 

[Noah Drakes]

and of the initial value of r(t) is 500

 

[Noah Drakes]

V= ∫_t0 ^t₁ (A/m) (r(t))^4/3

 

[haruspex]

Ok, I think i Got it
So the equation for velocity when acceleration varies from t₀ to t₁ is V=∫_t0 ^t1 (A/m) (r(t)) ^3/4

Yes and no.
That integral gives the increase in velocity in the time interval. If there is an initial velocity you would need to add that on. Strangely, the question does not specify an initial velocity, so unless you have omitted that in the problem statement I suppose you will have to take it as being from a standing start... very strange for a jet already in the air.
You are still leaving out the ".d" term, the one that specifies the variable of integration, as in ∫y.dx. You may think that is implied, but strictly speaking it is not.