As the speed time graph for a body thrown vertically upwards is not differentiable at the top point so why acc. is taken as 9.8 m/s² at the top point?

Question

In the shown speed - time graph for a body thrown vertically up there is a sharp kink in the graph at the instant for which body is at highest position ie not differentiable so we can't find acceleration. But at this highest point acceleration is due to rate of change of speed only as no angular velocity is there at highest point. Which means no radial acceleration component is there at highest point.

Now my question is using which logic do we say that value of acceleration at highest point is $9.8 ~\textrm{m/s}^2$? Does this situation shows that slope of speed - time graph may not represent magnitude of acceleration?

Answer 1

In physics speed is the term for the modulus of the velocity. Your graph is indeed not differentiable at the top of the trajectory, but that's OK because acceleration is the derivative of the velocity not the speed.

If we take the upwards direction to be positive then the velocity is positive when the object is moving upwards and negative when it is moving downwards. So the graph looks like this:

The velocity is differentiable at $v = 0$ so the acceleration is defined at that point.