A typical jetliner lands at a speed of 1.60 times 102 mi/h and decelerates at th

Question

A typical jetliner lands at a speed of 1.60 times 102 mi/h and decelerates at the rate of (10 mi/h)/s. If the plane travels at a constant speed of 1.60 times 102 mi/h for 1.00 s after landing before applying the brakes, what is the total displacement of the aircraft between touchdowm on the runway and coming to rest? 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. To find the displacement while braking, we could have used the two kinematics equations involving time, namely, delta x = v0t + 1/2 at2 and v = v0 + at, but because we werent interested in time, the time-independent equation was easier to use. By how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? m

Explanation / Answer

After converting into SI units, the initial velocity is 71.5 m/s

and, acceleration is -4.47 m/s2

Therfore, distance travelled by plane while coasting,

with a=0 ; u= 71.5

xcoasting = ut + 1/2 at2

              = 143 m

Distance travelled after the brakes are applied,

with a= -4.47 and u= 71.5

xbraking   = (v2 - u2)/2a

               = (0 - 71.52 )/2*(-4.47)

               = 572 m

Total distance travelled = 572 + 143 = 715 m

               Change in distance = 71.5 m,

(NOTE: This change in distace is only due to change in time of the coasting, which is increased by 71.5 m)

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