Answer:
If and only if the object has 0 speed. Any other case is a no.
Explanation:
In increasing math complexity:
If the graph of a distance/time is a straight line with a non-zero slope it means that the velocity-time graph is an horizontal line where every point has the same second coordinate, equal to the value of the slope of the first line (ie, if the first line has equation [tex]y=2t[/tex] the second line will have equation [tex]y=2[/tex] - I'm using y to denote either the distance or the speed, read it as "the non-time variable")
If the graph of a speed/time is a straight line, the graph of the distance/time is the area of the region of space beneath the straight line in the first graph, ie a triangle. And as you increase the sides of that triangle keeping the angles fixed, you get a parabolic curve.
Bonus case: if [tex]v(t)= 0; d(t) = 0[/tex], they are both straight lines, they both pass through the origin, and they are the exact same line.
With calculus (or: math becomes fun after you add the alphabet to it)
You are asking if the answer of the IVP
[tex]\left \{ {{x'=x} \atop {x(0)=0}} \right.[/tex]
is a linear function. The answer is: besides the obvious [tex]x(t)'=x(t)=0 \ \forall t[/tex] there are no other solutions, linear or otherwise.
