Acceleration, velocity and displacement

AI Thread Summary
The discussion revolves around how to graphically represent acceleration and velocity from the position equation x(t) = (at)i + (bt^2)j. The position graph yields a quadratic curve, and the velocity is derived as the first derivative of the position, while acceleration is the second derivative. To illustrate these vectors, one should plot the velocity vector v(t) and the acceleration vector a(t) at specific time points on the graph of position versus time. The key is to draw the vectors at the corresponding position x(t), ensuring their tails start at that point. The problem emphasizes the importance of differentiating the position equation to obtain the correct representations of velocity and acceleration.
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Homework Statement



Show the acceleration and velocity on the graph: x(t) = (at)i + (bt^2)j
(I have done the differentiation, but I need to show them physically on the graph)

a and b are positive constants

Homework Equations



x(t) = (at)i + (bt^2)j

The Attempt at a Solution



Well the graph is j against I
Hence, you get a quadratic, if you let at = x and bt^2 = y
you get to y is proportional to x^2 (b/a^2) is the constant of proportionality.

Now the velocity is the gradient (Right?) and the acceleration is the gradient, gradient (But because it has a local minimum we can say the acceleration is shown by this as the second derivative is positive = 2b j)

But how do i physically show the acceleration on the graph? Is it like circular motion? So it goes inwards or not?
 
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I'm not entirely sure what the intent of this problem is. It seems to be asking you to plot the velocity vector and the acceleration vector as a function of time. So after you differentiate the position vector \vec{x}(t) to get \vec{v}(t), you'll need to plot v_y(t) versus v_x(t).

It'll probably help to choose some representative values of t and plot the corresponding (v_x, v_y) points, and then you'll get a sense of the shape of the curve and you can just fill it in.

Then do the same thing for acceleration.
 
I think the problem just wants you to draw the vectors v and a on the graph of x. At some time t, the object will be at some point x(t). At that point, you'd assign the vectors v(t) and a(t), so draw them in with their tails at x(t).
 
It says "Sketch the path of the particle and indicate on your diagram a and v".

Currently i have the graph of i against j and it looks like y=x^2.

I don't understand where to put a or v (I thought v was the gradient)
 
No, v is the time derivative of the position, not the gradient.
 
the equation u have is a position equation, if graph it would be a graph of position v. time. once you have taken the derivative of the position equation you will have an equation with which you can graph velocity(v), giving you a graph of velocity v. time. then continue and take the derivative of the velocity equation this will give you the equation for acceleration with which you can use to graph acceleration v. time.

Hope this helps!
 
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