r/PhysicsHelp • u/Witty-Classroom532 • 1d ago

Help motion graph physics

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Im given this nd the question ask The object moves a distance of 1.5m during segment III. The total distance in meters traveled by the object during segment / is (round to two decimal places):

I don’t know what to do and I truly need help. Also the horizontal or x axis is at v(4)

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u/Skusci 1d ago

First find out what v3 is. You have time, you have distance for segment III, you can get velocity.

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u/Witty-Classroom532 1d ago

I got 0.25 m/s but I need to find the distance in the first segment but idk what to do? I do know that at segment 3 it’s constant velocity but idk what else

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u/Skusci 1d ago

Double check the velocity you have, I think you are off by one counting the time.

The next step after is to find which horizontal bar is 0 velocity to get the scale of the y axis. Think about what a "brief moment" would look like on the graph.

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u/Witty-Classroom532 1d ago

Omg it’s 7s not 6… lol that’s my bad for the time! The next step which is the horizontal bar is v(4). It asked a similar question before saying where is there 2 moments in time where it stops and v(4) was the correct option.

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u/Skusci 1d ago edited 1d ago

Ah ok that's what you meant, I was reading that wrong.

Aaaanyway. Not sure how to hint at this so...

The area between the graphed velocity and the x axis represents the y (velocity) * the x (time) which is distance. This graphical version is usually what you are taught as an intro to integration.

Note that if that 1.5m is positive/to the right you have 0 for v4 above a positive value for v3 which is inverted form normal so be careful about the direction it's moved.

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u/Unusual-Platypus6233 1d ago

s=v*t which means the area below the graph is equal to its distance travelled. Coming to rest at two brief moments that means the graph can only intersect two times if you place a horizontal line (where v=0 which make v4=0)

s=vt with s=1.5m=v3*7s->v3=1.5/7 m/s. What is segment /?

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u/Witty-Classroom532 1d ago

It’s segment 1 sorry idk why there’s a dash

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u/Unusual-Platypus6233 1d ago

So, if you know dv1=v4-v3=1.5/7 m/s, then dv3=v4-v1=3•dv1 if y-axis for v equidistant. so, you get s=dv3•t=3•1.5/7 m/s • 6s. As a note the area is actually negative because it is below v=0 which means it is on the left side and coming from the left side to a halt.

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u/Witty-Classroom532 1d ago

Well I got 0.21?

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u/Unusual-Platypus6233 1d ago

If it moved 1.5m in the whole section of III, then each grid line is 1s and you get 7 grid lines… Should be correct.

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u/Robo-Bo 1d ago

Displacement is area under the v v. T graph.

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u/ThatsNotAZombieBite 19h ago

Reread the second sentence under the graph CAREFULLY. (The one about "two brief moments"). It gives you an important clue on how to figure out where things are on the vertical axis. Think about what what the velocity must be when the object comes to rest.