• From ta = -25 s until to = -15 s the plane is taxiing east along the runway at a constant speed of 4 m/s. From ts until te = -10 s the plane slows down at a constant rate and comes to rest. • From te until to = 0 the plane turns around, turning nearly in place so that by to it is facing west. • At to = 0 the plane is near the east end of the runway, at rest. From to until ti = 5.0 s it speeds up with a constant acceleration going west. At t, it is going at a speed of 8.75 m/s. • After t, the plane continues speeding up, but at a steadily reducing rate (i.e. it is still speeding up, but the magnitude of its acceleration is decreasing). It lifts off at t2 = 65 s going at a speed of 85 m/s, having gone a distance of 2980 m down the runway. • From t2 until ta = 240 s the plane is climbing at an angle of 5.0° above horizontal, speeding up at a constant rate. At ta it is going at a speed of 230 m/s. • At ts the plane levels off, so it is flying at a constant height. From t3 until ta = 270 s it is following a circular path going at constant speed, so that by ta it has gone through 1/4 of a circle and is going due south.
• From ta = -25 s until to = -15 s the plane is taxiing east along the runway at a constant speed of 4 m/s. From ts until te = -10 s the plane slows down at a constant rate and comes to rest. • From te until to = 0 the plane turns around, turning nearly in place so that by to it is facing west. • At to = 0 the plane is near the east end of the runway, at rest. From to until ti = 5.0 s it speeds up with a constant acceleration going west. At t, it is going at a speed of 8.75 m/s. • After t, the plane continues speeding up, but at a steadily reducing rate (i.e. it is still speeding up, but the magnitude of its acceleration is decreasing). It lifts off at t2 = 65 s going at a speed of 85 m/s, having gone a distance of 2980 m down the runway. • From t2 until ta = 240 s the plane is climbing at an angle of 5.0° above horizontal, speeding up at a constant rate. At ta it is going at a speed of 230 m/s. • At ts the plane levels off, so it is flying at a constant height. From t3 until ta = 270 s it is following a circular path going at constant speed, so that by ta it has gone through 1/4 of a circle and is going due south.
Related questions
Question
Transcribed Image Text:• From ta = -25 s until to = -15 s the plane is taxiing east along the runway at a constant speed of 4 m/s. From ts
until te = -10 s the plane slows down at a constant rate and comes to rest.
• From te until to =0 the plane turns around, turning nearly in place so that by to it is facing west.
• At to = 0 the plane is near the east end of the runway, at rest. From to until ti = 5.0 s it speeds up with a constant
acceleration going west. At t, it is going at a speed of 8.75 m/s.
• After t, the plane continues speeding up, but at a steadily reducing rate (i.e. it is still speeding up, but the magnitude
of its acceleration is decreasing). It lifts off at t2 = 65 s going at a speed of 85 m/s, having gone a distance of 2980
m down the runway.
• From t2 until t3 = 240 s the plane is climbing at an angle of 5.0° above horizontal, speeding up at a constant rate.
At t3 it is going at a speed of 230 m/s.
• At t3 the plane levels off, so it is flying at a constant height. From t3 until ta = 270 s it is following a circular path
going at constant speed, so that by t4 it has gone through 1/4 of a circle and is going due south.
Transcribed Image Text:• From t4 until ts = 290 s the plane is speeding up at a constant rate, continuing its turn, on the same circular path
(same radius as from t3 to t4). At ts it is flying due east.
On this assignment we are going to look at the motion from t until t3. As on Assignment #2, we will use axes
so that the x-axis points west along the runway and the y-axis points south. The origin is at the east end of the
runway. At ta the plane is 80 m from the end of the runway, going east as described above. Note that the z-axis
points up. The entire motion from ta until t3 occurs in the xz-plane (i.e. y = 0 for this whole part of the motion).
As discussed in class, the unit vector that points along the z-axis is called k.
(a) Sketch x vs. t, v, vs. t, and a, vs. t graphs covering the time from t, until to. Your sketches do not need
to be numerically precise, but they should qualitatively show the behaviours of the plane during each time
interval. Label the times ta, t, ter to, ti, t2 on the time axis of the graphs. You should stack the graphs so
that their time axes line up and it is advisable to draw vertical lines through all three graphs at key times.
(b) Find the position of the plane at te. Don't forget to assess your answer.
(c) Find the position of the plane at t1. Assess...
(d) As the description of the motion says, the acceleration from ti until ta is not constant (it is decreasing in
magnitude). So the familiar equations you know (the UAM equations) do not apply to the time period
between t and to. Without more information there is nothing we can do to find the velocity or position
at any time between t, and t2. However, we can find averages. Find the average velocity, and average
acceleration between t, and t2. Assess...
(e) Suppose the x-component of acceleration between t and t2 is described by the function
a(t) a1- B(t-ti),
where a = 1.75 (in some units that you are about to think about) and B= 0.019 (in some units). What
are the units of a and B?
(f) Consider the time thatfway= (t+t2)/2 35 s. This is the time half way between ty and to. If the acceleration
was constant from ty until t2 then the specd at thatfway Wwould be (v + 2)/2 46.9 m/s (halfway between
the speeds at ty and t2). Given the function a(t) given above, is the speed v(thatfway) equal to, less than,
or greater than (v+ v2)/2? Explain.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
