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.

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:

(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 = (x_{f} - x_{i})/(t_{f}
- t_{i}) = (5 m - (-3m))/(6s - 1 s) = (8/5) m/s.

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.

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/s^{2} 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/s^{2})t.

We can construct the** table** below.

t | v_{x} |
---|---|

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.