Physics (calculus) homework help

[Shadow Conception]

This is the graph relevant to the question.

It's a position-time graph for a particle moving along the x-axis (one-dimensional motion).

The instruction is to determine the instantaneous velocity at t = 2.00 s by measuring the slope of the tangent line shown in the graph. The answer in the back of the book is -3.8 m/s.

However, slight problem. How the hell do you measure the slope when no actual points are given?

You could find the derivative of the quadratic y = (x-4)^2 + 2 is y = 2x + 8 and then plug in 2.00 to find -4 m/s. But according to the back of the book, that's not the answer. This could mean the formula I found for the quadratic isn't even accurate.

I also tried to estimate the point on the line at t = 2.00 s, and I get around 5.8 m. So I used the formula for velocity and got 5.8 m/2.00 s. This yields 2.9 m/s, which is still not the answer.

I am really at a loss right now. How do you do this problem? Thanks so much to anyone that helps!

 

[MikeMike]

uh, 13/3.5 = 3.71...

although that slope line doesnt exactly end at 13, but thats pretty damn close to the answer

 

[DrPizza]

You have a line? If so, find to points on the line, and find the slope between those points.

 

[Leros]

Lol, they gave you the x and y intercepts on the graph. So you have two points.

 

[DrPizza]

If you estimate that the y-intercept is 13.2, you end up with 3.77
If you estimate that the y-intercept is 13.3, you end up with exactly 3.8

 

[DrPizza]

Also, are you *sure* it's a quadratic? Just because it looks like one doesn't necessarily mean it is one.

(although, if you selected 3 points from that curve & came up with a 2nd degree polynomial for it, it's probably just as good a method for estimating the slope at a point.)

 

[Shadow Conception]

Wait what...

The slope of the given line is the instantaneous velocity? OH FUCK.

The slope of the line is tangent to the point we're talking about, which means the slope is the instantaneous velocity, right? Whereas, the derivative of the equation I found (assuming it was close to the truth) just graphs instantaneous velocities at points?

And I just guessed it was a quadratic because of how much like a parabola it looked. The vertex seems to be roughly at (4, 2), so I just built off of that in hopes that I would get an answer that would match the book.

Thanks for the quick answers, I think I get it now...

Edit: So what exactly did I find when I did 5.8 m/3.0 s? Average velocity? My own stupidity?

 

[Shadow Conception]

And one last question:

The position of a particle moving along the x-axis varies in time according to the expression x = 3t^2, where x is in meters and t is in seconds. Evaluate it's position at 3.00 s + delta t.

I can evaluate it's position at 3.00 s, but 3.00 s + delta T? What the hell? I don't even know where to start. Do I substitute t in the equation x = 3t^2 for 3.00 s + delta T, meaning x = 3(3.00 s + delta T)? ..and if so, then what?

Sorry I'm asking so many questions, our teacher did not give us any examples of these types of problems, and I'm only in Precalc, meaning my calculus skills are very weak right now.

 

[Whoozyerdaddy]

42

 

[dighn]

Originally posted by: Shadow Conception
And one last question:

The position of a particle moving along the x-axis varies in time according to the expression x = 3t^2, where x is in meters and t is in seconds. Evaluate it's position at 3.00 s + delta t.

I can evaluate it's position at 3.00 s, but 3.00 s + delta T? What the hell? I don't even know where to start. Do I substitute t in the equation x = 3t^2 for 3.00 s + delta T, meaning x = 3(3.00 s + delta T)? ..and if so, then what?

Sorry I'm asking so many questions, our teacher did not give us any examples of these types of problems, and I'm only in Precalc, meaning my calculus skills are very weak right now.

substitute (3 + DT) in place of t in the equation i.e. 3*(3 + DT)^2 . that's your answer. you got it almost right (missing the ^2)

 

[WildHorse]

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