Which of the graph lines corresponds to the relation y = k/x where k is a constant?
@ganeshie8
Suppose you are on a conveyor belt walkway at an airport. The walkway is moving at 0.50 m/s. You walk backward at 0.80 m/s to hand your suitcase to a friend who is 3.0 m behind you. A security guard stands alongside the walkway watching this. What is your speed relative to the guard? a. 0.30 m/s b. 0.50 m/s c.1.30 m/s d. 0.80 m/s
@ganeshie8 thank you so much for always helping me when I need it
I see two questions in this thread. Which question are you working on right now ?
I think is graph A for the first one and I just want a check , I picked A because is a hyperbole . And idk how to do the second one :/
graph A is correct for first q
thank you and how do I do the second one?
Suppose you are on a conveyor belt walkway at an airport. The walkway is moving at 0.50 m/s. You walk backward at 0.80 m/s to hand your suitcase to a friend who is 3.0 m behind you. A security guard stands alongside the walkway watching this. What is your speed relative to the guard?
-0.3?
Careful, the question is NOT about "velocity"
its just about the "speed"
It seems, you have the right idea. The speed with respect to the security guard is "0.3"
yes but that is where I get lost
velocity is a vector which requires both magnitude and direction speed is a scalar which has no direction
therefore would it stay at 0.8?
@ganeshie8
@Somy
I think your first thought is correct, the only thing is that its not '-0.3' but just 0.3m/s just like @ganeshie8 said.
basically saying since the walkways's speed is 0.5 m/s FORWARD, but you need to go backward a certain distance, first you need a speed that would overcome the walkway's speed + the speed you'd need to actually reach where you are going against the walkway's speed so taking a difference would give you the overall speed your total speed - walkways speed = your actual speed without walkway's since the guard is standing he has 0 speed so its only your speed that matters thats why i think 0.3 is correct choice im no pro in physics tho :3
alright thank you so much :)
no problem :)
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