Cesena plane

A 1043kg Cessna 172E (airplane) lands on a straight runway with the touchdown speed of  180 km/h, whereupon the brakes are applied causing a retarding force proportional to the velocity of the plane($k*V$). Find:

a) the expression for the deceleration of the plane,

b) the time required to reduce its velocity to 18 km/h after touchdown,

c) the distance traveled during that interval (units of $k=109.515$).


Force is retarding so is opposing plane motion.

$F =m*a$  that is $-k V = m*a$  or $a = -k V/m$


$d V/d t = -(K/m)*V$   or $d V/V =-(K/m)*d t$

$\int_{V0}^V (d V/V) = -(K/m)*\int_0^t d t$

$ln(V/V0) =-Kt/m$   that is $V =V0*exp(-Kt/m)$
$t = m/K*ln(V0/V) =1043/109.515*ln(180/18) = 21.93 sec$


$V =d x/d t = V0*exp(-(Kt)/m)$

$\int_0^d d x = V0*\int_0^t exp(-\frac{Kt}{m})d t$

$d = \frac{V_0*m}{K}[1-exp(-\frac{Kt}{m})]$

$d =(180*1000/3600)*1043/109.515*[1-exp(-\frac{109.515*21.93}{1043})]$

$d =428.57 m$