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Acceleration is regarded as a change in velocity, so what will be the acceleration of this ball in the figure, as its magnitude is constant, but the direction is changing.

enter image description here

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  • $\begingroup$ $a=dv/dt$ , i.e change in velocity / time duration $\endgroup$ – Koolman Dec 10 '17 at 11:06
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If the curved bit of your trajectory is an arc of a circle radius $R$ then the magnitude of the acceleration is $\frac{30^2}{R}$ - centripetal acceleration in appropriate units - and its direction is towards the centre of the circle.

If the trajectory is not an arc of a circle then one would have to evaluate the acceleration as a function of position both in terms of magnitude but also in terms of direction.
That evaluation of $\frac {d\vec v}{dt}$ might have to be done numerical rather than analytical.

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  • $\begingroup$ The trajectory is not an arc of a circle how would I, in this case, evaluate dv/dt ? $\endgroup$ – Sushant Dec 13 '17 at 12:17
  • $\begingroup$ What is the trajectory? $\endgroup$ – Farcher Dec 13 '17 at 12:18

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