Question 2.6: Runway Length GOAL Apply kinematics to horizontal motion wit...
Runway Length
GOAL Apply kinematics to horizontal motion with two phases.
PROBLEM A typical jetliner lands at a speed of 1.60 × 10² mi/h and decelerates at the rate of (10.0 mi/h)/s. If the plane travels at a constant speed of 1.60 × 10² mi/h for 1.00 s after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest?
STRATEGY See Figure 2.18. First, convert all quantities to SI units. The problem must be solved in two parts, or phases, corresponding to the initial coast after touchdown, followed by braking. Using the kinematic equations, find the displacement during each part and add the two displacements.
Convert units of speed and acceleration to SI:
v_0 = (1.60 × 10² \cancel{mi}/\cancel{h})(\frac{0.447 m/s}{1.00 \cancel{mi}/\cancel{h}}) = 71.5 m/s
a = (-10.0 (\cancel{mi}/\cancel{h})/s)(\frac{0.447 m/s}{1.00 \cancel{mi}/\cancel{h}}) = -4.47 m/s²
Taking a = 0, v_0 = 71.5 m/s, and t = 1.00 s, find the displacement while the plane is coasting:
Δx_{coasting} = v_0t + \frac{1}{2} at² = (71.5 m/s) (1.00 s) + 0 = 71.5 mUse the time-independent kinematic equation to find the displacement while the plane is braking.
v² = v_0² + 2aΔx_{braking}Take a = 24.47 m/s² and v_0 = 71.5 m/s. The negative sign on a means that the plane is slowing down.
Δx_{braking} = \frac{v² – v_0 ²}{ 2a} = \frac{0 – (71.5 m/s)²}{ 2.00(-4.47 m/s²)} = 572 mSum the two results to find the total displacement:
Δx_{coasting} + Δx_{braking} = 71.5 m + 572 m = 644 mREMARKS To find the displacement while braking, we could have used the two kinematics equations involving time, namely, Δx = v_0t + \frac{1}{2} at² and v = v_0 + at, but because we weren’t interested in time, the time-independent equation was easier to use.