# Constructing Antiderivatives

1. Nov 25, 2007

1. The problem statement, all variables and given/known data
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?

3. 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.

2. Nov 25, 2007

### CompuChip

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. Nov 25, 2007

### HallsofIvy

Staff Emeritus
??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.

No, the initial speed is NOT 200 mph- that's the final speed. It starts at 0 mph.

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. Nov 25, 2007

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.