Answer to Question #139783 in Physics for Aaron

Question

A typical jetliner lands at a speed of 71.5 m/s. If the plane travels at a constant speed of 71.5 m/s for 1.00 s, then apply the brakes and decelerates at a rate of -4.47 m/s2. What is the total displacement of the aircraft between touchdown and coming to rest?

Answer 1

The distance covered when the aircraft was travelling with the speed $v_0 = 71.5m/s$ for $t_1 = 1s$ is:

$$d_1 = v_0t_1 = 71.5m/s\cdot 1s = 71.5m$$

Time it took the aircraft to stop from $v_0 = 75.1m/s$ to $v = 0$ under the acceleration $a = -4.47m/s^2$ is:

$$t_2 = \dfrac{v_0}{|a|}$$

The distance travelled in time $t_2$ is:

$$d_2 = v_0t_2 - \dfrac{|a|t_2^2}{2} = \dfrac{v_0^2}{|a|} - \dfrac{v_0^2}{2|a|} = \dfrac{v_0^2}{2|a|}$$

Substituting numbers, obtain:

$$d_2 = \dfrac{v_0^2}{2|a|} = \dfrac{71.5^2}{2\cdot 4.47} \approx 571.84m$$

The total distance:

$$d = d_1 + d_2 = 643.34m$$

Answer. 643.34 m.