Question

The following is the distance-times table of an object in motion:
Time in seconds Distance in metres
$0$ $0$
$1$ $1$
$2$ $8$
$3$ $27$
$4$ $64$
$5$ $125$
$6$ $216$
$7$ $343$
$\left(a\right)$ What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
$\left(b\right)$ What do you infer about the forces acting on the object?

Answer 1

According to the data given, we conclude that the distance covered by an object is directly proportional to the cube of time taken.

$\therefore$ $x\propto t^3$ $⟹x=k{t}^3$ where $k$ is constant

Differentiating w.r.t time, $\frac{dx}{dt}=k\left(3t^2\right)$

Differentiating again w.r.t time, $\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left(3k{t}^2\right)=6kt$

$⟹$ Acceleration of the object $\frac{d^{2}x}{d{t}^2}=a=6kt$

Hence, the acceleration of the object is not constant and increases with time.

Force acting on the object $F=ma=m\left(6kt\right)$

$⟹F=6kmt$ i.e force acting on the object increases with time.