Motion with constant velocity

Let us assume that a cart is moving with constant speed of 2 m/s in the positive x-direction and that at t = 0 it passes through the origin. 
We can represent this motion in various ways.
We may use a formula and write

v = ∆x/∆t,  x_f - x_i = v*(t_f - t_i).

If we choose our coordinate system so the cart is at position x = 0 at time t = 0, then x(t) = v*t.
We say that the position x increases linearly with the time t,  x = (2 m/s)*t.
We can construct the table below.

We may also represent the motion using a position versus time graph or a velocity versus time graph.  A position versus time for our cart is shown below.
The instantaneous velocity v(t) = ∆x/∆t as ∆t becomes infinitesimally small is equal to the slope of the position versus time graph at time t.

For motion with uniform velocity in one dimension the position versus time graph is a straight line.  The slope ∆x/∆t of this straight line is equal to v.  The velocity versus time graph yields a straight line with zero slope.  A velocity versus time graph for our cart is shown below.

Problem:

At t = 1 s, a particle moving with constant velocity is located at x = -3 m, and at t = 6 s the particle is located at x = 5 m.
(a) From this information, plot the position as a function of time.
(b) Determine the velocity of the particle from the slope of this graph.

Solution:

• Reasoning:
  For motion with uniform velocity in one dimension the position versus time graph is a straight line.  The slope ∆x/∆t of this straight line is equal to v.
• Details:
  (a) When the velocity is constant, the instantaneous velocity is equal to the average velocity. 
  
  (b) The average velocity of the particle is
  v = (x_f - x_i)/(t_f - t_i) = (5 m - (-3m))/(6s - 1 s) = (8/5) m/s.

Motion with non-uniform velocity

When an object moving in one dimension is accelerating, then

• the position versus time graph is NOT a straight line,
• the velocity versus time graph is NOT a straight line with zero slope.

External links:
Position versus time graphs
Velocity versus time graphs

Motion with constant acceleration in one dimension

Let us assume at t = 0 a cart is leaving the origin with zero initial velocity and constant acceleration of 2 m/s² in the positive x-direction.  Then

a = ∆v/∆t,  v_f - v_i = a*(t_f - t_i).

If the cart has velocity v = 0 at time t=0, then

v(t) = a*t.

We may say that v increases linearly with the time t, v= (2 m/s²)t.
We can construct the table below.

A velocity versus time graph for the cart is shown below.
The instantaneous acceleration a(t) = ∆v/∆t as ∆t becomes infinitesimally small is equal to the slope of the velocity versus time graph at time t.

For motion with constant acceleration in one dimension the velocity versus time graph is a straight line.  The slope of this straight line yields a. 
The acceleration versus time graph yields a straight line with zero slope.

Note:  When representing motion using a graph, make sure you label the axes properly.  When presented with a motion graph first look at the axes to identify what is plotted.

Summary:

We can represent one-dimensional motion using a position versus time graph or a velocity versus time graph.

Position versus time graph:
The slope gives the instantaneous velocity. (v_x = lim_∆t-->0∆x/∆t )

Positive slope --> positive velocity
Negative slope --> negative velocity
Constant slope --> constant velocity
Changing slope --> acceleration
(The position versus time graph of motion with constant acceleration is a section of a parabola.)

Velocity versus time graph:
The slope gives the instantaneous acceleration. (a_x = lim_∆t-->0∆v_x/∆t )
Positive slope --> positive acceleration
Negative slope --> negative acceleration
Constant slope --> constant acceleration