Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

(A) Determine the displacement of the particle in the time intervals t = 0 to t

ID: 2992275 • Letter: #

Question

(A) Determine the displacement of the particle in the time intervals t = 0 to t = 1 s and t = 1 s to t = 3 s.

From the graph, form a mental representation of the motion of the particle. Keep in mind that the particle does not move in a curved path in space such as that shown by the brown curve in the graphical representation. The particle moves only along the x axis in one dimension. At t = 0, is it moving to the right or to the left?

During the first time interval, the slope is negative and hence the average velocity is negative. Therefore, we know that the displacement between and must be a negative number having units of meters. Similarly, we expect the displacement between and to be positive.
In the first time interval, set ti = tA = 0 and tf = tB = 1 s and use the following equation to find the displacement:
?xA?B = xf - xi = xB - xA
?xA?B = [ -4(1) + 2(1)2 ] - [ -4(0) + 2(0)2 ]
?xA?B =

m
For the second time interval (t = 1 s to t = 3 s), set ti = tB = 1 s and tf = tD = 3 s:
?xB?D = xf - xi = xD - xB
?xB?D = [ -4(3) + 2(3)2 ] - [ -4(1) + 2(1)2 ]
?xB?D =

m
These displacements can also be read directly from the position-time graph.
(B) Calculate the average velocity during these two time intervals.
In the first time interval, use the equation for average velocity with ?xt = tf - ti = tB - tA = 1 s:
vx, avg (A ?B)=
?xA?B
?xt
=
-2 m
1 s
vx, avg (A ?B) =

m/s
In the second time interval, ?xt = 2 s:
vx, avg (B ?D)=
?xB?D
?xt
=
8 m
2 s
vx, avg (B ?D) =

m/s
These values are the same as the slopes of the lines joining these points in the figure.
(C) Find the instantaneous velocity of the particle at t = 2.5 s.
Measure the slope of the green line at t = 2.5 s (point ) in the figure: vx= m/s
Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected?

NOTE: just answer what is vx= m/s

Explanation / Answer

Hi, there is no graph in the question. Please post the same so that the question can be answered. Thanks.