Q A1.     The following is the distance-time table of an object in motion:                  Time in seconds  Distance in metre 0 0 1 1 2 8 3 27 4 64 5 125 6 216 7 343     (a)     What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

From the table, the relation between time and the distance can be seen.

$s\ =\ t^3$

Thus the velocity of the particle is increasing with time.

$v\ =\ \frac{ds}{dt}\ =\ 3t^2$

and                                                               $a\ =\ \frac{dv}{dt}\ =\ 6t$

Hence acceleration increases linearly with time.

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