How to calculate instantaneous acceleration if -->

[Hardikph]

I tried but I can't.

 

[Ray Vickson]

View attachment 102497
I tried but I can't.View attachment 102497

You must show your work. What have you tried?

 

[Hardikph]

You must show your work. What have you tried?

First I used the formula dv/dt but I am used to have a ƒ(x) in the questions I usually did.

 

[haruspex]

First I used the formula dv/dt but I am used to have a ƒ(x) in the questions I usually did.

Write dv/dt in terms of x and y, as defined in this question.
What is the gradient at a point on a graph, a) in terms of algebra, and b) in terms of geometry?

 

[CWatters]

+1

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

 

[Hardikph]

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

 

[Nidum]

Edit : Distorted plot corrected .

 

[Nidum]

 

[rcgldr]

There's a conflict in the figure, t2 is stated as being 3.8, but shows as 4.0 in the figure. There's insufficient information to determine instantaneous acceleration without more sample points or a limitation on the formula for acceleration versus time. If the figure was more accurate, the multiple sample points could be used to generate a function for position versus time, and then velocity and acceleration versus time. For example, the graph could be a hyperbola, in which case acceleration approaches zero as time and distance increase, or the graph could be a parabola, in which case acceleration is constant, although the graph seems to be closer to being a hyperbola.

 

[CWatters]

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.

 

[rcgldr]

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.

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.

 

[haruspex]

@Hardikph , I don't know if you've been able to follow the discussion between CWatters and rcgldr.
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?

 

[Hardikph]

Now I Understand. Thank You guys.