Velocity/Acceleration

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victorias

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at t =2, isn't velocity constant (slope)? Then how can there be an acceleration?

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Velocity is not constant. It is always changing in this case because it's a sine curve. At each infinitesimal moment in time, the ball's velocity is changing. In other words, dv/dt does not equal zero. Imagine tossing a ball up into the air. You would apply the classical equations of motion to describe the mechanics of the ball, correct? Would you say that the ball's acceleration is zero at the top of its climb? Or imagine dropping a ball from a window high above the ground. At the moment of release, the ball has zero velocity. But does it experience acceleration?

What's important is the change in velocity, not the absolute velocity, i.e. dv/dt.
 
Actually at that point the acceleration is at a maximum no?

Yes: F = -k*x = m*a. Therefore, a = -k*x/m. k and m are constants for a particular case and thus acceleration depends only on displacement of spring from equilibrium and obviously the crests and troughs of the function represent the maximum displacements.
 
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