Constructing Antiderivatives

In summary, the 727 jet needs to accelerate from 0 to 200 mph in 30 seconds. If the runway is long enough, the jet will reach its final speed.
  • #1
ada0713
45
0

Homework Statement


A 727 jet needs to attain a speed of 200 mph to take off. It it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be?

The Attempt at a Solution


First I converted mi/hr to mi/sec
200mi/hr * 1hr/60min * 1min/60sec = 0.056 mi/sec
and since accerelation is 0.045
v(t) =0.056t +C

and since the initial speed is 200mph (=0.045mps)
v(0)= 0+C= 0.056
v(t) = 0.056t + 0.056

then I integrated it s(t)=0.028t^2 +0.056t+C
but what do I do next?
I am stuck on finding the distance because I don't know how to find the constant.
 
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  • #2
What about calling the start of the runway s = 0?
Basically, it's exactly the same question as here, so I'll expand a bit on my answer there.
 
  • #3
ada0713 said:

Homework Statement


A 727 jet needs to attain a speed of 200 mph to take off. It it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be?

The Attempt at a Solution


First I converted mi/hr to mi/sec
200mi/hr * 1hr/60min * 1min/60sec = 0.056 mi/sec
and since accerelation is 0.045
v(t) =0.056t +C
??The acceleration is not "0.045" (I don't know where you got that figure!) nor is it 0.056 mi/sec because that is the final speed. If it goes from 0 to 200mi/hr= 0.056 mi/sec in 30 sec then it's acceleration is 0.056/30 m/sec2.

and since the initial speed is 200mph (=0.045mps)
v(0)= 0+C= 0.056
v(t) = 0.056t + 0.056
No, the initial speed is NOT 200 mph- that's the final speed. It starts at 0 mph.

then I integrated it s(t)=0.028t^2 +0.056t+C
but what do I do next?
I am stuck on finding the distance because I don't know how to find the constant.

Once you have the correct coefficients, you evaluate that at t= 30 seconds. (You were told it needs to accelerate to 200 mph and it can do that in 30 seconds.) Don't worry about the constant. Either take the initial position to be 0 or, since you are only interested in how far the jet goes, subtract its postion at t= 0 (which will be just "C") from it's position at t= 30- which means the "C" cancels out.

By the way, there's a much easier way to do this problem. If something accelerates at a constant rate, it "average speed" is just the average of the initial and final speeds. Here the initial speed is 0 and the final speed is 200 mph. What was its average speed? How far will it go in 30 seconds at that average speed?
 
  • #4
0.045 was a typo. sorry about that:)
so,,
Vo=0, Vf=0.056, and a=0.056/30=0.0019.

then, I set
V(t)=0.0019t+Vo
V(30)=0.0019*30+Vo=0.056
therefore Vo=0 (I guess I didn't even have to plug 30 back in t since
the initial velocity has to be zero anyway since it would've started from the rest,,)

so V(t)= 0.0019t
and s(t)=0.00095t^2+C

s(30)-s(0)= (0.855+C)-C

therefore runway has to be 0.855 miles long.
Does the answer seem right?
 

1. What is an antiderivative?

An antiderivative is the inverse operation of a derivative. It is a function that, when differentiated, gives back the original function. In simpler terms, it is the "undoing" of a derivative.

2. How do you construct an antiderivative?

To construct an antiderivative, you must first identify the original function's derivative. Then, you can use the power rule, product rule, quotient rule, and chain rule to work backwards and find the antiderivative. Additionally, you can use known antiderivatives of basic functions to construct the antiderivative of more complex functions.

3. Can any function have an antiderivative?

No, not all functions have antiderivatives. A function must be continuous and differentiable on its entire domain in order for it to have an antiderivative. If a function has a discontinuity or is not differentiable at any point, it does not have an antiderivative.

4. What is the relationship between a derivative and an antiderivative?

The derivative and antiderivative are inverse operations of each other. This means that if you take the derivative of a function and then find the antiderivative of the derivative, you will get back the original function. Symbolically, this can be represented as ∫(f'(x))dx = f(x) + C, where C is the constant of integration.

5. Can multiple antiderivatives exist for the same function?

Yes, multiple antiderivatives can exist for the same function. This is because when finding an antiderivative, you are essentially finding a family of functions that all have the same derivative. This family of functions can differ by a constant, represented by the constant of integration. Therefore, there can be an infinite number of antiderivatives for a single function.

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