The acceleration-velocity (a vs v) graph of a moving particle is shown. Four points P, Q, R, S are marked on the graph. The particle is

speeding up at P
speeding up at Q
speeding up at S
speeding down at R

Correct answer: speeding up at S

Solution

Speeding up requires a and v to have the same sign (a*v > 0). In the standard version of this graph, S lies where both a and v are negative, so a*v > 0 and the particle is speeding up at S.

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