The figure shows the position of a particle m | vedclass

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(A) The velocity of the particle is given by the slope of the position-time graph,$v = \frac{dx}{dt}$.$1$. The particle comes to rest when the velocit

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The particle has come to rest $4$ times.

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(A) The velocity of the particle is given by the slope of the position-time graph,$v = \frac{dx}{dt}$.$1$. The particle comes to rest when the velocity is zero,which corresponds to the points where the slope of the $x-t$ graph is zero (the peaks and troughs of the curve).$2$. Looking at the graph,there are peaks at approximately $t \approx 3.5 \, s$ and $t \approx 8.5 \, s$,and troughs at approximately $t \approx 1 \, s$ and $t \approx 6 \, s$. Thus,the particle comes to rest at $4$ distinct points.$3$. At $t=8 \, s$,the graph is sloping upwards,meaning the velocity is positive.$4$. Between $t=2 \, s$ and $t=6 \, s$,the graph goes from a peak to a trough,meaning the slope is negative (velocity is negative).$5$. Since the slope is changing,the velocity is not constant.Therefore,the correct statement is that the particle has come to rest $4$ times.

The figure shows the position of a particle moving on the $x$-axis as a function of time.

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