An airplane starts from rest and accelerates at a constant rate of 3.0 m/s. It r

Question

An airplane starts from rest and accelerates at a constant rate of 3.0 m/s. It reaches its takeoff after traveling a distance of 4800 ft along the runway on a day with no wind. Answer the following questions, showing your work: (a) How fast is the plane going after accelerating for 8 seconds ? Give the answer both in m's and in miles per hour. (b) What is the distance traveled by the plane in the first 8 seconds of its run ? (c) What is the time interval from when the plane starts ad (d) What is the (horizontal) speed of the plane as it just lifts off the ground, i.e. its takeoff speed Give your answer both in m/s and in miles per hour. Now suppose that the same plane takes off on a different day when there is a headwind of 6.0 m/s That means that if the plane's current speed along the ground is v., its speed moving through the air-L i. e. its "airspeed"-is (v, + 6.0 m/s). Because the lift provided by the wings depends on airspeed, the plane takes off when its airspeed reaches the takeoff speed you found in part (d). (e) How much does that headwind change the time interval (compared to part () over which the plan needs to accelerate for it to take off? That is, does the plane take off sooner, or later, and by how many seconds? (D By how much does that headwind change the distance the plane needs to travel along the runway in order to take off?

Explanation / Answer

4800 ft = 1463.04 m

c) using: s= ut + 0.5at2

1463.04 = 0*t + 0.5*3*t2

t= 31.230 seconds

d) using: v2 - u2 = 2as

v2 - 0 =2*3*1463.04   {u=0 since it starts at rest}

v= 93.692 m/s OR v=209.583 mi/hr

e) new takeoff speed, v=93.692 - 6 = 87.692 m/s

using: v=u+at

t = v/a {u=0 since it starts at rest}

t=87.692/3; t= 29.230 seconds

The plane can take off 3 seconds sooner than earlier

f) using: s=ut+0.5at2

s=0+0.5*3*29.2302

s=1281.589 m

change in distance = (1463.04-1281.589) = 181.451 m

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