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(D) Maximum acceleration corresponds to the steepest slope of the $v - t$ graph,which represents the maximum rate of change of velocity.$1$. For the i

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$6$

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(D) Maximum acceleration corresponds to the steepest slope of the $v - t$ graph,which represents the maximum rate of change of velocity.$1$. For the interval $t = 0$ to $t = 20 \, sec$: $a = \frac{20 - 0}{20 - 0} = 1 \, cm/sec^2$.$2$. For the interval $t = 20$ to $t = 30 \, sec$: $a = 0 \, cm/sec^2$ (constant velocity).$3$. For the interval $t = 30$ to $t = 40 \, sec$: $a = \frac{80 - 20}{40 - 30} = \frac{60}{10} = 6 \, cm/sec^2$.$4$. For the interval $t = 40$ to $t = 80 \, sec$: $a = \frac{0 - 80}{80 - 40} = \frac{-80}{40} = -2 \, cm/sec^2$.Comparing the magnitudes of acceleration in all intervals,the maximum acceleration is $6 \, cm/sec^2$. Thus,the correct option is $D$.

The $v - t$ graph of a moving object is given in the figure. The maximum acceleration is .......... $cm/sec^{2}$.

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