Question

How does the graph of displacement vs time look for something moving at a constant, positive velocity?

Answer 1

Straight line with slope `=` constant velocity. See sample graph below.

Explanation:

General expression for displacement /time is
`s=s_@+ut+1/2at^2` .........(1)
`s` is displacement, `s_@ and u` is initial displacement and velocity respectively, `t` being the time and `a` is the acceleration.

In the given problem, the object is moving at a constant, positive velocity, suppose `v`.

Now first derivative of velocity with respect to time `(dv)/dt` is acceleration `a` which is `=0`.
Therefore, equation (1) becomes

`s=vt+s_@` .....(2)
Compare with the standard equation for a straight line
`y=mx+c`

In our equation (2), `v` is the slope of the line and `c=s_@`, i.e., `y` intercept is `s_@`.
For example, if constant velocity `=0.5ms^-1`, and `s_@=5m`, the graph will look like

graph{y=0.5x+5 [-12.45, 7.55, -0.04, 9.96]}