# Differential Equation Problem Jet Take Off

• Dec 1st 2008, 09:03 PM
ishanj07
Differential Equation Problem Jet Take Off
I really can't figure out this problem nor can i figure a similar problem from an example in class.

Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 45 seconds, how long must the runway be(in feet)? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.

Thanks!
• Dec 1st 2008, 10:58 PM
Quote:

Originally Posted by ishanj07
I really can't figure out this problem nor can i figure a similar problem from an example in class.

Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 45 seconds, how long must the runway be(in feet)? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.

Thanks!

$
\frac{d^2y}{dt^2}= a
$

$
$

$
y=\int{atdt}=at^2/2
$

hence
$
y= \frac{(220-0)*22/15}{45} * {45}^2= ans
$

• Dec 2nd 2008, 12:17 AM
mr fantastic
Quote:

$
\frac{d^2y}{dt^2}= a
$

$
$

But dy/dt = 0 when t = 0 therefore C = 0.

$
y=\int{atdt}=at^2/2 {\color{red}+ D}
$

But y = 0 when t = 0 therefore D = 0.

hence
$
y= \frac{(220-0)*22/15}{45} * {45}^2= ans
$

Mr F edits in red.

I'm sure you did this in your head but best to spell it out .... the arbitrary constant often gets forgotten - a disaster when it's not equal to zero (Wink)
• Dec 2nd 2008, 09:20 AM
Wazdinero
Don't forget to divide the answer given above by 2 which comes when you take the anti derivative of the acceleration*time in the second integral.

If you don't, the answer you get above will be 14520 feet, which is incorrect.
(at^2) = 14520 (INCORRECT!)

Divide by 2 and you will have the correct answer:
(at^2)/2 = 7260 (CORRECT!)