Question

an airplane lands and starts down the runway at a northwest velocity of 55.7 m/s. what constant acceleration allows it to come to a stop in 1.24 km?

Answer 1

Let's choose the northwest direction as positive direction of x-axis and choose the initial point of coordinate system at landing point of the airplane. Using the laws for uniformly accelerated motion ($l$ - travelled distance along the x-axis, ${v_0}$ - initial speed of airplane).

$$l = {v_0}t + \frac{{a{t^2}}}{2}$$

$$v = {v_0} + at$$

Using the second equation we can get the time needed for particular change of speed

$$t = \frac{{v - {v_0}}}{a}$$

Substituting this to the first equation

$$l = {v_0}\frac{{v - {v_0}}}{a} + \frac{a}{2}{(\frac{{v - {v_0}}}{a})^2}$$

and expanding the brackets and simplifying the expression

$$l = \frac{{{v_0}v}}{a} - \frac{{v_0^2}}{a} + \frac{a}{2}\frac{{{v^2} - 2v{v_0} + v_0^2}}{{{a^2}}} = \frac{{{v_0}v}}{a} - \frac{{v_0^2}}{a} + \frac{{{v^2}}}{{2a}} - \frac{{v{v_0}}}{a} + \frac{{v_0^2}}{{2a}} = \frac{{{v^2} - v_0^2}}{{2a}}$$

Expressing the acceleration

$$a = \frac{{{v^2} - v_0^2}}{{2l}}$$

Also we know that airplane stopped after travelling 1.24 km, then $v=0$. Now let's do the calculations

$$a = - \frac{{v_0^2}}{{2l}} = - \frac{{{{55.7}^2}[\frac{{{m^2}}}{{{s^2}}}]}}{{2 \cdot 1.24[km]}} = - \frac{{3102.49[\frac{{{m^2}}}{{{s^2}}}]}}{{2 \cdot 1.24 \cdot {{10}^3}[m]}} = - 1.251[\frac{m}{{{s^2}}}]$$

Minus sign means that the vector of acceleration directed opposite to the vector of initial velocity, that is in our case the vector of acceleration directed southeast.