In circular motion, say a ball is moving at a constant speed. The direction of the ball is changing so the velocity of the ball is changing to the ball is accelerating. Even though it's moving at a constant speed. Is no one else baffled by that? I mean, how would you plot that on a velocity-time graph? Can someone clarify?

Question

michele_laino · Answer

every time we have a change of the vector velocity, we have an accelerated motion. So it is in circular motion, where the vector velocity is changing during the motion of the particle. Here is the situation:
|dw:1431875587214:dw|
the change of the vector velocity, is the vector v_b - v_A. So we have an acceleration. We can demonstrate, that such vector difference is along the radius of the circumference, and it is directed towards the center of the circumference

anonymous · Answer

So how would that be plotted on a velocity-time graph?

michele_laino · Answer

That graph is like the graph of the rectilinear uniform motion of a particle

anonymous · Answer

I am only at GCSE level, so I don't really know what you're talking about.

michele_laino · Answer

If I call with s the space traveled by our particle on the circumference, and I call with t time needed to travel that space, then we can write:
s=v*t. So Using the coordinates s, and t, as cartesian coordinates, we have, the subsequent graph:
|dw:1431878884620:dw|

michele_laino · Answer

where s_0 is the space traveled by our particle, from the origin of spaces, at time t=0

michele_laino · Answer

more precisely, we have to write the subsequent formula:
$$\Large  s = {s_0} + vt$$

anonymous · Answer

And what does v represent?

anonymous · Answer

Wait, that's a stupid question, never mind.

michele_laino · Answer

v is the speed (constant value) of our particle, or, in other words, v represents the length of each tangent vector of my draving above

michele_laino · Answer

and s is the length of the arc of circumference