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Member advised to use the homework template for posts in the homework sections of PF.

**I tried but I can't.
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- Thread starter Hardikph
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- #1

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Member advised to use the homework template for posts in the homework sections of PF.

- #2

Ray Vickson

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You must show your work. What have you tried?

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First I used the formula dv/dt but I am used to have a ƒ(x) in the questions I usually did.You must show your work. What have you tried?

- #4

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Write dv/dt in terms of x and y, as defined in this question.First I used the formula dv/dt but I am used to have a ƒ(x) in the questions I usually did.

What is the gradient at a point on a graph, a) in terms of algebra, and b) in terms of geometry?

- #5

CWatters

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+1

It looks like this question is testing that you know what dx/dt and dv/dt mean in the real world.

It looks like this question is testing that you know what dx/dt and dv/dt mean in the real world.

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I am NOT getting it. Can you be a little more specific.

- #7

Nidum

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Edit : Distorted plot corrected .

Last edited:

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Nidum

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- #9

rcgldr

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- #10

CWatters

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There's a conflict in the figure, t2 is stated as being 3.8, but shows as 4.0 in the figure.

Sorry I disagree. I think that's deliberate.

It states the velocity at t=3.8 seconds is 130m/s and asks you to calculate the acceleration at t2=4.0 seconds. The graph gives you enough data to calculate the velocity at t2=4.0. So you have the velocities at two points a known time apart.

The accuracy might not be great but you can calculate an answer.

- #11

rcgldr

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I wasn't paying attention to the limited choices of possible answers, so the accuracy doesn't need to be that great. There's no information after t = 4.0, so what's being calculated is the average acceleration from t = 3.8 to t = 4.0, not the instantaneous acceleration at t = 4.0. Based on the graph and the data, the acceleration is decreasing with time, instantaneous acceleration at t = 4.0 would be less.It states the velocity at t=3.8 seconds is 130m/s and asks you to calculate the acceleration at t2=4.0 seconds. The graph gives you enough data to calculate the velocity at t2=4.0. So you have the velocities at two points a known time apart.

The accuracy might not be great but you can calculate an answer.

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You are told the velocity at t=3.8s, and there is enough information in the graph to find it at t=4s. Can you see how to deduce that second velocity? Having got the two velocities, can you see how to estimate the acceleration?

- #13

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Now I Understand. Thank You guys.

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