1. The problem statement, all variables and given/known data
You are consulting for an elevator company. During a safety test, technicians created the following graph of the velocity of the elevator versus time; you have been asked to interpret the graph. Unfortunately, you are told that the technician who printed the graph has made two mistakes. The labeled interval on the velocity axis should be 2.5 m/s rather than 2.0 m/s, and the labeled time interval should be 3.6 s rather than 4.0 s. You should begin by redrawing the graph on your own paper with the proper labels on the velocity and time axes.
Since this motion is purely one-dimensional, give your answers to the following questions using a coordinate system consisting of a y-axis in the Up-Down direction with Up as positive. Indicate directions by the appropriate plus or minus sign. Assume that the test begins (at time zero) with the elevator at position y0=-3.2 m (i.e. 3.2 meters below ground level -- ground level has been selected as the location at which y=0).
(a) What is the position of the elevator at time 1.8 seconds?
(b) What is the position of the elevator at time 7.2 seconds?
2. Relevant equations
##X_f = x_0 + V_x(ΔT)##
Maybe ##v = d/ΔT##
##v_x = Δx/Δt##
3. The attempt at a solution
Attempt 1:
Just plug everything in, ##v_x = v_xavg = v## so ##v_x = 1.389 m/s##
##x_f = -3.2m + 1.389m/s(1.8s)##
## x_f = -0.7m## WRONG!!
Attempt 2:
Tried to multiply by the velocity at time 1.8s according to my redrawn graph, which was 2.5
##x_f = -3.2m + 2.5m/s * 1.8s##
##x_f = 1.3m## WRONG ONCE AGAIN!!
Attempt 3:
Tried to plug my numbers into a similar problem someone asked about on this forum
##v_x = -3.2 + (2.5/1.8) * 1.8## I don't even know but it was wrong again.
I just would like some guidance on what I need to be looking at because obviously my way is wrong.
I'm going to attach the original graph, and my redrawn graph.
You are consulting for an elevator company. During a safety test, technicians created the following graph of the velocity of the elevator versus time; you have been asked to interpret the graph. Unfortunately, you are told that the technician who printed the graph has made two mistakes. The labeled interval on the velocity axis should be 2.5 m/s rather than 2.0 m/s, and the labeled time interval should be 3.6 s rather than 4.0 s. You should begin by redrawing the graph on your own paper with the proper labels on the velocity and time axes.
Since this motion is purely one-dimensional, give your answers to the following questions using a coordinate system consisting of a y-axis in the Up-Down direction with Up as positive. Indicate directions by the appropriate plus or minus sign. Assume that the test begins (at time zero) with the elevator at position y0=-3.2 m (i.e. 3.2 meters below ground level -- ground level has been selected as the location at which y=0).
(a) What is the position of the elevator at time 1.8 seconds?
(b) What is the position of the elevator at time 7.2 seconds?
2. Relevant equations
##X_f = x_0 + V_x(ΔT)##
Maybe ##v = d/ΔT##
##v_x = Δx/Δt##
3. The attempt at a solution
Attempt 1:
Just plug everything in, ##v_x = v_xavg = v## so ##v_x = 1.389 m/s##
##x_f = -3.2m + 1.389m/s(1.8s)##
## x_f = -0.7m## WRONG!!
Attempt 2:
Tried to multiply by the velocity at time 1.8s according to my redrawn graph, which was 2.5
##x_f = -3.2m + 2.5m/s * 1.8s##
##x_f = 1.3m## WRONG ONCE AGAIN!!
Attempt 3:
Tried to plug my numbers into a similar problem someone asked about on this forum
##v_x = -3.2 + (2.5/1.8) * 1.8## I don't even know but it was wrong again.
I just would like some guidance on what I need to be looking at because obviously my way is wrong.
I'm going to attach the original graph, and my redrawn graph.
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