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.


ticker 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.


vector diagram

Link:  Vector Diagrams

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.


image

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.


image

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.
image
(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

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,  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.


image

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.