Answer
(a) $v = 179~mph$
(b) The acceleration is constant during takeoff.
(c) t = 34.8 s
(d) The A320 can not take off safely on a 2.5-mile long runway.
Work Step by Step
(a) $v = (80~m/s)(\frac{3600~s}{1~hour})(\frac{1~mi}{1609~m})$
$v = 179~mph$
(b) From t = 0 to t = 10 s:
$a = \frac{\Delta v}{\Delta t} = \frac{23~m/s}{10~s}$
$a = 2.3~m/s^2$
From t = 10 to t = 20 s:
$a = \frac{\Delta v}{\Delta t} = \frac{23~m/s}{10~s}$
$a = 2.3~m/s^2$
From t = 20 to t = 30 s:
$a = \frac{\Delta v}{\Delta t} = \frac{23~m/s}{10~s}$
$a = 2.3~m/s^2$
The acceleration is constant during takeoff.
(c) $t = \frac{\Delta v}{a} = \frac{80~m/s}{2.3~m/s^2}$
$t = 34.8~s$
(d) We can find the takeoff distance $x$
$x = \frac{v^2-v_0^2}{2a}$
$x = \frac{(80~m/s)^2-0}{(2)(2.3~m/s^2)}$
$x = 1391.3~m$
Three times the takeoff distance is 4173.9 meters. We can convert this distance to miles.
$x = (4173.9~m)(\frac{1~mi}{1609~m})$
$x = 2.6~miles$
The A320 can not take off safely on a 2.5-mile long runway because three times the takeoff distance is 2.6 miles.
