## Representing motion

### 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,  xf - xi = v*(tf - ti).

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.

Time (t) Position (x)
1 s 2 m
2 s 4 m
3 s 6 m
4 s 8 m
5 s 10 m

Diagrams can also represent the motion.  One type of motion diagram is a ticker-tape diagram.  A stationary "ticker" makes a mark at regular time intervals.  A moving object drags a tape past the ticker, and a trail of marks is left on the tape.  For an object moving with uniform velocity the marks are spaced uniformly on the tape.

Link:  Ticker Tape Diagrams

A vector diagram can also describe the motion of our cart.  At evenly-spaced time intervals we depict the relative magnitude and direction of a vector quantity, such as the velocity.  For an object moving with uniform velocity, the vectors all have the same length and the same direction.

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) = dx/dt 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 dx/dt 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:
(a) When the velocity is constant, the instantaneous velocity is equal to the average velocity.
The position versus time graph is a straight line, and the slope of this line is equal to v.

(b) The average velocity of the particle is
v = (xf - xi)/(tf - ti) = (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

• on a ticker tape diagram the marks are NOT spaced uniformly,
• in a vector diagram the velocity vectors do NOT all have the same length and direction,
• the position versus time graph is NOT a straight line,
• the velocity versus time graph is NOT a straight line with zero slope.

### 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,  vf - vi = a*(tf - ti).

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.

t vx
1 s 2 m/s
2 s 4 m/s
3 s 6m/s
4s 8 m/s
5 s 10 m/s

A velocity versus time graph for the cart is shown below.
The instantaneous acceleration a(t) = ddv/dt 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.