Newton's Laws Conceptual Problem

AI Thread Summary
Thrust reversers in commercial aircraft redirect ejected air to enhance performance during landing and takeoff. The discussion highlights that directing air forward can slow the plane down, while downward thrust can enable vertical takeoff without a runway. The user initially selected options B, C, and D as correct, but faced feedback indicating an error, particularly regarding option C, which may have been misinterpreted by the problem writer. The conversation suggests that the phrasing in the problem could lead to confusion about acceleration and deceleration. Overall, thrust reversers are critical for improving aircraft maneuverability and safety during specific flight phases.
JSmithDawg
Messages
9
Reaction score
0

Homework Statement


"Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrust reversers that can change the direction of the ejected air, sending it forward."

"When might thrust reversers be useful in practice? (Chose all that apply)
a) If the ejected air is directed forward then thrust force is backward. This might be desirable to conserve the speed of the plane.
b) If the ejected air is directed downward, the thrust force is up. Jets can take off vertically without needing a runway this way.
c) If the ejected air is directed forward then thrust force is backward. This might be desirable to accelerate the plane.
d) If the ejected air is directed forward then thrust force is backward. This might be desirable to slow the plane down. "

The Attempt at a Solution


A - No. If thrust is moving a plane forward, then it reversing the direction of the thrust should slow it down rather than conserve the speed
B - Yes. The thrust should allow the jet to rise even though it's not on a runway
C - Yes. This will cause the plane to change velocity, which'll change acceleration
D - Yes. If a plan. If thrust is moving the plane forward, then reversing the thrust will cause the plane to slow down.

My Answer: B,C,D, but the website told me I was wrong and didn't provide me any feedback. What am I doing wrong?
 
Physics news on Phys.org
The problem writer probably wasn't as clever as you when considering C. In some situations the desire might well be to slow the plane's forward speed (a negative acceleration is still an acceleration). The problem writer may have (mistakenly) considered the phrase "desirable to accelerate the plane" to mean only speed it up in the forward direction. This is an error in the problem, not an error on your part. So, it's likely that your "C = True" answer is your culprit. Your other choices look okay to me.
 
I would agree with "C = True" if it weren't for this (bold by me). "... but commercial aircraft often have thrust ..."

Actually using it while in flight would probably have similar results as:

_503780_reverse_thrust5.gif


Thrust reversers can use a variety of safety locks to prevent the system from accidentally deploying, including sensors and switches to check that the aircraft is actually landing at a low speed.
http://news.bbc.co.uk/2/hi/science/nature/503780.stm
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top