Displacement vs. Time Graph

[chancemd]

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

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, 0s, 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?

---

Members don't see this ad.

 

[imbackasd]

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

 

[chancemd]

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

thank you thank you!!!🙂 I overlooked that big time, gots to be more careful!!!

 

[chancemd]

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

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

and wont't it be -2m/s since it has a downward slope/moving in opposite direction?


I also meant to ask, for 50 s how did they get the particle travelling 40m with a displacement of 0m?

I apologize for all the ?'s im just trying to make sure b/c in my practice im always getting these types of ?'s wrong....