Question

An airplane starts from rest, travels 5000ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. it then climbs in a straight line with a uniform acceleration of 3 ft/s^s until it reaches a constant speed of 220 mi/h. draw the st, vt, and at graphs that describe the motion.

Answer 1

Answer:

Explanation:

Given

Take off speed $v=162\ mph\approx 237.6\ ft/s$

distance traveled in runway $d=5000 ft$

using motion of equation

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(237.6)^2=2\times a\times 5000$

$a=5.64\ ft/s^2$

Acceleration after take off $a_2=3\ ft/s^2$

time taken to reach $237.6 ft/s$

$v=u+at$

$237.6=0+5.64\times t$

$t=42.127\ s$

after take off it take $t_2$ time to reach $220 mph\approx 322.67$

$322.67=237.6+3\times t_2$

$t_2=28.35\ s$

total time taken $t_0=t+t_1$

$t_0=70.48\ s$