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.
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.
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.
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:
When an object moving in one dimension is accelerating, then
Links:
Position versus time graphs
Velocity versus time graphs
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) = dv/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.
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. (vx = 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. (ax = lim∆t-->0∆vx/∆t
)
Positive slope --> positive acceleration
Negative slope --> negative acceleration
Constant slope --> constant acceleration