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Speed time graph

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?

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  • $\begingroup$ 9.8m/s/s is for the graph of the component of velocity that points towards earth. The speed is different from that. $\endgroup$
    – doublefelix
    Apr 6 at 7:24
  • $\begingroup$ @MichaelSeifert sure, good point. I've deleted my comment. $\endgroup$
    – Marius Ladegård Meyer
    Apr 6 at 14:11

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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:

Velocity:time graph

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

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  • $\begingroup$ Another example showing the different between acceleration and the rate of change of speed: an object in uniform circular motion is constantly accelerating, but its speed is constant and the slope of its speed graph would be zero. $\endgroup$
    – Michael Seifert
    Apr 6 at 11:48

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