HELP!! Displacement vs. Time Graph and Velocity vs. Time GraphEK Physics

[chancemd]

Hello everyone, I need a little help! My question stems from EK Physics Pg 8-9 btw!



*I already posted this in the MCAT Discussions,but I felt like it'll be better if I post it here...

This was my original post:

I'm not understanding how on the Displacement vs. Time Graph for 50 sec the particle is moving 2m/s to the Left. Now I understand the that the particle is back where it started, 0m, and I understand why its its moving Left (b/c it has a downward slope, therefore a neg velocity) but im not understanding how they got 2 m/s. Are they using one of the Linear Motion Equ to get this value? Idk im confused?

Im also confused on how the Av. Velocity of the particle after 100s is 20m/100sec, again i understand why its to the right, but its the magnitude im having trouble understanding? Again are they using the Linear Motion Equ to get these values?

The one person that replied said this:

Slope of the graph, Displacement was 20 meters and the time took 10 seconds. Thus 20/10 = 2 m/s

Also Avg velocity = displacement / time. Displacement = 20 meters after 100 seconds. Thus 20/100

I initially thought I understood their reply, but when I went back I got confused again. 🙁

Where did they get the 10s from? Is it from 10s as in the time it took get to 50s 4rm the last point, which was 40 s?

Also to add, at 50s, how did they get a Distance of 40m and a Displacement of 0? I think I understand how they got it, I just want to double check.

And for the Av Velocity, im still kind of confused on that one too, so please exaplain.

Basically, for the response i received above, Im not understanding how they got the displacement and time values for it. Im not understanding the magnitudes? *Hopefully Im making sense with where my confusion lies*


So to add onto my ?'s, as for the Velocity vs. Time Graph (EK Pg. 9),
-at 10s, how did they get 50m to the Right?
-at 20s, how is the total displacement 0m to the Right and traveled a distance of 100m
-I understand whats going on btw 20s-40s! Thank God! LOL! 🙂
-at 80s, how is the particle decelerating to the right (or accelerating to the Lt) I kind of understand this, but again I just need someone to clarify.
-at 100s, I understand how to do this b/c of Chad! 🙂

PLEASE HELP YA'LL, I KEEP GETTING THESE TYPES OF QUESTIONS WRONG ON PRACTICE PASSAGES I DO!! 🙁​

OH AND PLEASE EXPLAIN HOW YOU GOT YOUR DISPLACEMENT OR TIME ANSWERS B/C IM UTTTERLY CONFUSED ANY HELP IS DEF APPRECIATED!!! THANKS SO MUCH!!!
​

---

Members don't see this ad.

 

[ingramw1202]

Take a picture of the graph and post it please

 

[umdnjmed]

Hello everyone, I need a little help! My question stems from EK Physics Pg 8-9 btw!



*I already posted this in the MCAT Discussions,but I felt like it'll be better if I post it here...

This was my original post:

I'm not understanding how on the Displacement vs. Time Graph for 50 sec the particle is moving 2m/s to the Left. Now I understand the that the particle is back where it started, 0m, and I understand why its its moving Left (b/c it has a downward slope, therefore a neg velocity) but im not understanding how they got 2 m/s. Are they using one of the Linear Motion Equ to get this value? Idk im confused?

Im also confused on how the Av. Velocity of the particle after 100s is 20m/100sec, again i understand why its to the right, but its the magnitude im having trouble understanding? Again are they using the Linear Motion Equ to get these values?

The one person that replied said this:

Slope of the graph, Displacement was 20 meters and the time took 10 seconds. Thus 20/10 = 2 m/s

Also Avg velocity = displacement / time. Displacement = 20 meters after 100 seconds. Thus 20/100

I initially thought I understood their reply, but when I went back I got confused again. 🙁

Where did they get the 10s from? Is it from 10s as in the time it took get to 50s 4rm the last point, which was 40 s?

Also to add, at 50s, how did they get a Distance of 40m and a Displacement of 0? I think I understand how they got it, I just want to double check.

And for the Av Velocity, im still kind of confused on that one too, so please exaplain.

Basically, for the response i received above, Im not understanding how they got the displacement and time values for it. Im not understanding the magnitudes? *Hopefully Im making sense with where my confusion lies*


So to add onto my ?'s, as for the Velocity vs. Time Graph (EK Pg. 9),
-at 10s, how did they get 50m to the Right?
-at 20s, how is the total displacement 0m to the Right and traveled a distance of 100m
-I understand whats going on btw 20s-40s! Thank God! LOL! 🙂
-at 80s, how is the particle decelerating to the right (or accelerating to the Lt) I kind of understand this, but again I just need someone to clarify.
-at 100s, I understand how to do this b/c of Chad! 🙂

PLEASE HELP YA'LL, I KEEP GETTING THESE TYPES OF QUESTIONS WRONG ON PRACTICE PASSAGES I DO!! 🙁​

OH AND PLEASE EXPLAIN HOW YOU GOT YOUR DISPLACEMENT OR TIME ANSWERS B/C IM UTTTERLY CONFUSED ANY HELP IS DEF APPRECIATED!!! THANKS SO MUCH!!!
​

"Where did they get the 10s from? Is it from 10s as in the time it took get to 50s 4rm the last point, which was 40 s? "
Yes, the10 is from 50-40. Initially, the particle starts at position zero. Then it moves 20m to the right (i believe the positive y direction was defined as right). This takes 20s. Therefore according to Vavg = displacement/ time, the average velocity at the 20m point is 20/20 = 1m/s. Because no acceleration occurred (straight line as opposed to curve), this velocity must have been constant. Therefore from the get go, until the 20m mark, the particle had a constant velocity of 1m/s.

At the 20m mark, displacement was constant (horizontal line) Velocity is a CHANGE in displacement with time, implying that since there was no change in displacement, velocity was zero (the particle was motionless for 20 seconds [40-20]) Keep in mind that the slope of the line represents velocity. After 40s, the slope becomes negative (slopes downward), implying that the direction of the velocity changed from right to left. Thus the particle turned around and headed back for the origin. After 10s (50-40) it traveled 20m; Vavg = 20/10 = 2m/s. Again because of the straight line (no acceleration), this velocity was constant throughout the trip. Since the particle traveled 20m, turned around and traveled another 20m, the distance was 20+20 = 40m but the displacement was 0 since displacement is a change in position and the particle returned to the origin.

Each time the particle gets back to the x axis, it is back at the origin. At 100s, the displacement is 20m because at 97s, the particle had come back to the origin (change in position =0, so no displacement), and then in moved 20m to the right. Thus Vavg = 20m/100s

"-at 10s, how did they get 50m to the Right?"
The area between the curve an the x axis in a velocity vs time graph is equal to distance. Between zero and 10s, the area between the curve and x axis is that of a triangle. To find x, use 1/2 base times height = 1/2 (10)(10) = 50m.

"-at 20s, how is the total displacement 0m to the Right and traveled a distance of 100m"
Between 0 and 20s, you have two triangles, each has an area of 50m. 50 +50 = 100m. This is the distance traveled. Crossing the x axis in this plot means the particle turned around and started moving in the opposite direction. Therefore it traveled 50m in one direction (say left for example) then stopped, turned around and traveled 50m to the right, back to the origin. therefore displacement is zero.

"-at 80s, how is the particle decelerating to the right (or accelerating to the Lt) I kind of understand this, but again I just need someone to clarify."
The slope of the graph is equal to the acceleration of the particle. Also, For any segment above the x axis, the particle is moving to the right while for any segment below the x axis, the particle is moving to the left. in the first 10s for instance, the particle's velocity is decreasing, (it is slowing down) but it is moving in the "right direction." At 10s, after traveling 50m, it turns around and starts moving in the left direction and is speeding up because the magnitude of the velocity is actually increasing.

At 80s, the slope of the curve is negative. However, the velocity is positive (above the x axis). Because "above x axis" = "right," the slope, which is decreasing must be pointing towards the left. Because velocity and acceleration are in opposite directions, right and left respectively, the particle must be accelerating to the left or decelerating to the right.

I hope this helps. These graphs really frustrated me when I first started studying. When I finally understood them, my self confidence got a well deserved boost. I hope this helps. Good luck.

 

[mseclectic]

"Where did they get the 10s from? Is it from 10s as in the time it took get to 50s 4rm the last point, which was 40 s? "
Yes, the10 is from 50-40. Initially, the particle starts at position zero. Then it moves 20m to the right (i believe the positive y direction was defined as right). This takes 20s. Therefore according to Vavg = displacement/ time, the average velocity at the 20m point is 20/20 = 1m/s. Because no acceleration occurred (straight line as opposed to curve), this velocity must have been constant. Therefore from the get go, until the 20m mark, the particle had a constant velocity of 1m/s.

At the 20m mark, displacement was constant (horizontal line) Velocity is a CHANGE in displacement with time, implying that since there was no change in displacement, velocity was zero (the particle was motionless for 20 seconds [40-20]) Keep in mind that the slope of the line represents velocity. After 40s, the slope becomes negative (slopes downward), implying that the direction of the velocity changed from right to left. Thus the particle turned around and headed back for the origin. After 10s (50-40) it traveled 20m; Vavg = 20/10 = 2m/s. Again because of the straight line (no acceleration), this velocity was constant throughout the trip. Since the particle traveled 20m, turned around and traveled another 20m, the distance was 20+20 = 40m but the displacement was 0 since displacement is a change in position and the particle returned to the origin.

Each time the particle gets back to the x axis, it is back at the origin. At 100s, the displacement is 20m because at 97s, the particle had come back to the origin (change in position =0, so no displacement), and then in moved 20m to the right. Thus Vavg = 20m/100s

"-at 10s, how did they get 50m to the Right?"
The area between the curve an the x axis in a velocity vs time graph is equal to distance. Between zero and 10s, the area between the curve and x axis is that of a triangle. To find x, use 1/2 base times height = 1/2 (10)(10) = 50m.

"-at 20s, how is the total displacement 0m to the Right and traveled a distance of 100m"
Between 0 and 20s, you have two triangles, each has an area of 50m. 50 +50 = 100m. This is the distance traveled. Crossing the x axis in this plot means the particle turned around and started moving in the opposite direction. Therefore it traveled 50m in one direction (say left for example) then stopped, turned around and traveled 50m to the right, back to the origin. therefore displacement is zero.

"-at 80s, how is the particle decelerating to the right (or accelerating to the Lt) I kind of understand this, but again I just need someone to clarify."
The slope of the graph is equal to the acceleration of the particle. Also, For any segment above the x axis, the particle is moving to the right while for any segment below the x axis, the particle is moving to the left. in the first 10s for instance, the particle's velocity is decreasing, (it is slowing down) but it is moving in the "right direction." At 10s, after traveling 50m, it turns around and starts moving in the left direction and is speeding up because the magnitude of the velocity is actually increasing.

At 80s, the slope of the curve is negative. However, the velocity is positive (above the x axis). Because "above x axis" = "right," the slope, which is decreasing must be pointing towards the left. Because velocity and acceleration are in opposite directions, right and left respectively, the particle must be accelerating to the left or decelerating to the right.

I hope this helps. These graphs really frustrated me when I first started studying. When I finally understood them, my self confidence got a well deserved boost. I hope this helps. Good luck.

Thanks a lot for that clarification umdnjmed! I was having the same problem as chancemd in interpreting the graphs.