Question

The following is the distance-time table of an object in motion:

Time in seconds	Distance in meters
$0$	$0$
$1$	$1$
$2$	$8$
$3$	$27$
$4$	$64$
$5$	$125$
$6$	$216$
$7$	$343$

(A) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(B) What do you infer about the forces acting on the object?

Answer 1

Hint
In the question, we have the distance moved by an object in equal intervals of time. We have to find the behavior of the acceleration of that body. We know that acceleration is the rate of change of velocity with time and velocity is the rate of change of distance with time.
$\Rightarrow v = \dfrac{{{x_2} - {x_1}}}{t}$ (Where, ${\text{ }}v\;$is the velocity of the object,${\text{ }}{x_{2{\text{ }}}}$ is the final distance, and ${\text{ }}{x_1}{\text{ }}$is the initial velocity)
$\Rightarrow a = \dfrac{{v - u}}{t}$ (Where ${\text{ }}a{\text{ }}$stands for the acceleration of the object, ${\text{ }}v{\text{ }}$stands for the final velocity of the object, and ${\text{ }}u{\text{ }}$stands for the initial velocity of the object)

Complete step by step answer
We have to find the velocity and acceleration for each interval
For the first interval, the velocity is given by
$\Rightarrow {v_1} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{1 - 0}}{{1 - 0}} = 1{\text{ }}m/s$
For the second interval, the velocity is given by
$\Rightarrow {v_2} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{8 - 1}}{{2 - 1}} = 7{\text{ }}m/s$
The acceleration is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{7 - 1}}{1} = 6{\text{ }}m/{s^2}$
The velocity in the third interval is given by,
$\Rightarrow {v_3} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{27 - 8}}{{3 - 2}} = 19{\text{ }}m/s$
The acceleration of the interval is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{19 - 7}}{1} = 12{\text{ }}m/{s^2}$
The velocity of the fourth interval is given by
$\Rightarrow {v_4} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{64 - 27}}{{4 - 3}} = 37{\text{ }}m/s$
The acceleration is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{37 - 19}}{1} = 18{\text{ }}m/{s^2}$
The velocity of the fifth interval is given by
$\Rightarrow {v_5} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{125 - 64}}{{5 - 4}} = 61{\text{ }}m/s$
The acceleration is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{61 - 37}}{1} = 24{\text{ }}m/{s^2}$
The velocity of the sixth interval is given by
$\Rightarrow {v_6} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{216 - 125}}{{6 - 5}} = 91{\text{ }}m/s$
The acceleration is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{91 - 61}}{1} = 30{\text{ }}m/{s^2}$
The velocity of the seventh interval is given by,
$\Rightarrow {v_7} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{343 - 216}}{{7 - 6}} = 127{\text{ }}m/s$
The acceleration is given by,
$\Rightarrow a = \dfrac{{v - u}}{t} = \dfrac{{127 - 91}}{1} = 36{\text{ }}m/{s^2}$
From the calculation, we can see that the velocity is increasing as well as the acceleration is also increasing.
In conclusion, we can say that the acceleration is not uniform. We can see that the acceleration is increasing in each interval.
From Newton's second law of motion, we know that force is the product of mass and acceleration. We have inferred that the acceleration is increasing. Therefore the force acting on the body is also increasing.

Note
If the velocity of an object changes in either magnitude or direction, then we can say that the object is in accelerated motion. An object is said to be moving with uniform acceleration if the velocity changes by equal amounts in equal intervals of time, however small the intervals are.