Suppose that the position vector function for a particle is given by
**r**(t)
= x(t)**i **+ y(t)**j**

with x(t) = at + b and y(t) = ct^{2 }+ d, where a = 1 m/s, b = 1 m, c =
0.125 m/s^{2}, and d = 1 m.

(a) Calculate the average velocity during the time interval t = 2 s to t = 4 s.

(b) Determine the velocity and speed at t = 2 s.

Solution:

(a) The position of the particle is given to us as a function of time. At t =
2 s the position of the particle is

**r**(2 s) = [(1 m/s)(2 s) + 1 m]**i** + [(0.125 m/s^{2})(4 s^{2})
+ 1 m]**j **= 3 m** i **+ 1.5 m** j**.

At t = 4 s its position is

**r**(4 s) = [(1 m/s)(4 s) + 1 m]**i** + [(0.125 m/s^{2})(16 s^{2})^{
}+ 1 m]**j **= 5 m** i **+ 3 m** j**.

The average velocity of the particle between 2 and 4 seconds is

**v **= ∆**r**/∆t = (**r**(4 s) -
**r**(2 s))/(2 s),

**v **= [(5 m - 3 m)**i** + (3 m - 1.5 m)**j**]/(2 s) = (1 m/s)**i **
+ (0.75 m/s)**j**.

(b) The instantaneous velocity of the particle is

**v **= d**r**/dt = [d(x(t)/dt]**i **+ [dy(t)/dt]**j **= a**i **+
2ct**j**.

Review derivatives if you
don't understand this.

At t = 2 s the instantaneous velocity is **v**(2s) = (1 m/s)**i **+ (0.5
m/s)**j**.

The speed at 2 seconds is v = (v_{x}^{2} + v_{y}^{2})^{½}
= (1 + 0.25)^{½} m/s = 1.12 m/s.

The coordinates of an object moving in the xy-plane vary with time according
to the equations

x = (-5 m) sin(t) and y = (4 m) - (5 m) cos(t), where t is in seconds.

(a) Determine the components of velocity and components of acceleration at t =
0.

(b) Write expressions for the position vector, the velocity vector, and the
acceleration vector at any time t > 0.

(c) Describe the path of the object in an xy-plot.

Solution:

(a) The x- and y-components of the position vector of the object are given as a
function of time.

We have v_{x }= dx/dt = (-5 m/s) cos(t), v_{y }= dy/dt = (5 m/s)
sin(t).

At t = 0 we have **v **= (-5 m/s)**i**.

Differentiating the velocity vector with respect to time we find

a_{x }= dv_{x}/dt = (5 m/s^{2}) sin(t), a_{y }=
dv_{y}/dt = (5 m/s^{2}) cos(t).

At t = 0 we have **a** = (5 m/s^{2})**j**.

(b) **v **= (-5 m/s) cos(t)**i **+ (5 m/s) sin(t)**j**,
**a **= (5
m/s^{2}) sin(t)**i **+ (5 m/s^{2}) cos(t)**j**.

(c) A spreadsheet can be used to construct the table below.

t (s) | x(t) (m) | y(t) (m) |
---|---|---|

0 | 0 | -1 |

0.1 | -0.49917 | -0.97502 |

0.2 | -0.99335 | -0.90033 |

0.3 | -1.4776 | -0.77668 |

0.4 | -1.94709 | -0.6053 |

0.5 | -2.39713 | -0.38791 |

0.6 | -2.82321 | -0.12668 |

0.7 | -3.22109 | 0.175789 |

0.8 | -3.58678 | 0.516466 |

0.9 | -3.91663 | 0.89195 |

1 | -4.20735 | 1.298488 |

1.1 | -4.45604 | 1.732019 |

... | ... | ... |

We can now use the spreadsheet to make a Graph [chart type XY (Scatter) in Microsoft Excel] of the path of the object, by plotting y(t) as a function of x(t). The path is a circle. The center of this circle lies on the y-axis at x = 0, y = 4 m.

At t = 0 the particle is at x = 0, y = -1m. Its velocity vector is pointing into the negative x-direction. The particle is moving clockwise in a circle.

Let** a **= a_{x}**i **+ a_{y}**j **+ a_{z}**k**
= (a_{x}, a_{y}, a_{z}) = constant. Since
**a** is
constant, the components a_{x}, a_{y}, and a_{z} are
constant and the average acceleration is equal to the instantaneous
acceleration.

Assume that at t = 0 a particle is at position **r**_{0} = x_{0}**i
**+ y_{0}**j **+ z_{0}**k** and has velocity
**v**_{0}
= v_{0x}**i **+ v_{0y}**j **+ v_{0z}**k**. At
time t, i.e. after a time interval ∆t = t, its velocity has changed by an amount
∆**v **= **a**∆t = **a**t. We can rewrite this in terms of the
components as

∆v_{x}**i **+ ∆v_{y}**j **+ ∆v_{z}**k
**= a_{x}t** i **+ a_{y}t** j **+ a_{z}t** k**.

A vector equation like this is equivalent to a set of three equations, one for each component of the vector in three dimensions.

∆v_{x
}= a_{x}∆t, ∆v_{y }= a_{y}∆t, ∆v_{z }=
a_{z}∆t.

The x-component of the acceleration only changes the x-component of the
velocity, the y-component of the acceleration only changes the y-component of
the velocity, etc.

The velocity as a function of time is given by

v_{x }= v_{0x }+ ∆v_{x
}= v_{0x }+ a_{x}∆t,

v_{y }= v_{0y
}+ a_{y}∆t,

v_{z}
= v_{0z}∆t + a_{z}∆t,

or
**v **=
**v**_{0
}+ **a**∆t.

Note: If the directions of **v**_{0} and
**
a** are different, the direction of **v** is different from the direction
of **v**_{0}.

For constant acceleration the motions along the perpendicular axes
of a Cartesian coordinate system are independent and can be
analyzed separately.

The position of the particle at time t is given by

x = x_{0 }+ v_{0x}∆t + ½a_{x}∆t^{2},

y = y_{0
}+ v_{0y}∆t + ½a_{y}∆t^{2},

z = z_{0
}+ v_{0z}∆t + ½a_{z}∆t^{2},

or
** r **= **r**_{0 }+
**v**_{0}∆t + ½**a**∆t^{2}.

Note: The directions of **r**_{0},
**v**_{0},
**a**, and **r** may all be different. If
**a** is constant then the x, y, and z-coordinated as a function
of time can be found independently.

At t = 0, a particle moving in the xy-plane with constant acceleration has a
velocity **v**_{0 }= (3**i **- 2**j**) m/s at the origin. At t
= 3 s, the particle's velocity is **v** = (9**i **+ 7**j**) m/s. Find

(a) the acceleration of the particle and

(b) the coordinates at any time.

Solution:

(a) We are told that the particle moves with constant acceleration and we are
given its velocity vector at t = 0 and t = 3 s.

**a **= ∆**v**/∆t = (**v**(3 s) -
**v**_{0})/(3 s) = (9 m/s
- 3 m/s)**i**/(3 s) + (7 m/s + 2 m/s)**j**/(3 s).

a** **= (2 m/s^{2})**i **+ (3 m/s^{2})**j**.

(b) At t = 0 the particle is at the origin, **r**_{0 }= 0.
Therefore **r **= **v**_{0}t + ½**a**t^{2} is its
position at time t.

**r **= [(3 m/s)t + (1 m/s^{2})t^{2}]**i **+ [(2 m/s)t +
(1.5 m/s^{2})t^{2}]**j**.

x(t) = (3 m/s)t + (1 m/s^{2})t^{2}, y(t) = (2 m/s)t + (1.5 m/s^{2})t^{2}.

A particle originally located at the origin has
an acceleration of **a **= 3**j** m/s^{2} and an initial velocity
of **v**_{0 }= 5**i** m/s.

(a) Find the vector position and velocity at any time t.

(b) The coordinates and speed of the particle at t = 2 s.

Solution:

(a) A particle is moving with constant acceleration. At t = 0 it is at the
origin and its velocity **v**_{0} is given. Its position at any time
t is **r **= **v**_{0}t + ½**a**t^{2}_{.}

**r **= (5 m/s)t**i **+ (1.5 m/s^{2})t^{2}**j**.

The particles velocity at time t is **v **= **v**_{0 }+
**a**t.

**v **= (5 m/s)**i **+ (3 m/s^{2})t**j.**(b) At t = 2 s the coordinates of the particle are x = 10 m and y = 6 m.
Its speed is

v = (v