Question

learn more remarks to find the displacement while braking, we could have used the two kinematics equations involving time, namely, $delta x = v_0t+\frac{1}{2}at^{2}$ and $v = v_0 + at$, but because we werent interested in time, the time - independent equation was easier to use. question by how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? m practice it use the worked example above to help you solve this problem. a typical jetliner lands at a speed of 158 mi/h and decelerates at the rate of (11.7 mi/h)/s. if the jetliner travels at a constant speed of 158 mi/h for 1.3 s after landing before applying the brakes, what is the total displacement of the jetliner between touchdown on the runway and coming to rest? m exercise hints: getting started | im stuck! a jet lands at 11.3 m/s, the pilot applying the brakes 2.24 s after landing. find the acceleration needed to stop the jet within 5.95×10² m after touchdown. m/s² resources read it

Explanation:

Step1: Convert speed units for Practice It

First, convert 158 mi/h to m/s. 1 mi = 1609.34 m, 1 h = 3600 s. So $v_0=158\times\frac{1609.34}{3600}\text{ m/s}\approx70.6\text{ m/s}$.

Step2: Calculate displacement during constant - speed phase for Practice It

The jetliner travels at a constant speed $v_0$ for $t_1 = 1.3\text{ s}$. Using $x_1=v_0t_1$, we get $x_1 = 70.6\times1.3=91.78\text{ m}$.

Step3: Calculate displacement during braking phase for Practice It

We know $v = v_0+at$, when it stops $v = 0$. The deceleration $a=- 11.7\text{ mi/h/s}$. Convert it to m/s²: $a=-11.7\times\frac{1609.34}{3600}\text{ m/s²}\approx - 5.23\text{ m/s²}$. Using $v^{2}-v_{0}^{2}=2ax_2$, $0 - v_{0}^{2}=2ax_2$, so $x_2=\frac{-v_{0}^{2}}{2a}=\frac{-70.6^{2}}{2\times(- 5.23)}\text{ m}\approx477.7\text{ m}$.

Step4: Calculate total displacement for Practice It

The total displacement $x=x_1 + x_2=91.78+477.7 = 569.48\text{ m}$.

Step5: For the Exercise

The jet lands with $v_0 = 11.3\text{ m/s}$, it travels at constant speed for $t_1=2.24\text{ s}$, so the displacement during this phase $x_1=v_0t_1=11.3\times2.24 = 25.312\text{ m}$.
Let the acceleration be $a$, and the remaining displacement $x_2=595 - 25.312=569.688\text{ m}$, and the final velocity $v = 0$. Using $v^{2}-v_{0}^{2}=2ax_2$, $0-(11.3)^{2}=2a\times569.688$. Then $a=\frac{-11.3^{2}}{2\times569.688}\text{ m/s²}\approx - 0.112\text{ m/s²}$.

Answer:

Practice It: 569.48
Exercise: - 0.112