## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(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.
(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.