## Motion in 2 and 3 dimensions

### Let us review the definitions of velocity and acceleration with the help of two problems.

#### Problem:

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) = ct2 + d, where a = 1 m/s, b = 1 m, c = 0.125 m/s2, 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/s2)(4 s2) + 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/s2)(16 s2) + 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 = dr/dt = [d(x(t)/dt]i + [dy(t)/dt]j = ai + 2ctj.
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 = (vx2 + vy2)½ = (1 + 0.25)½ m/s = 1.12 m/s.

#### Problem:

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 vx = dx/dt = (-5 m/s) cos(t), vy = 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
ax = dvx/dt = (5 m/s2) sin(t), ay = dvy/dt = (5 m/s2) cos(t).
At t = 0 we have a = (5 m/s2)j.
(b)  v = (-5 m/s) cos(t)i + (5 m/s) sin(t)j, a = (5 m/s2) sin(t)i + (5 m/s2) 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 us now consider three-dimensional motion with constant acceleration.

Let a = axi + ayj + azk = (ax, ay, az) = constant.  Since a is constant, the components ax, ay, and az are constant and the average acceleration is equal to the instantaneous acceleration.
Assume that at t = 0 a particle is at position r0 = x0i + y0j + z0k and has velocity v0 = v0xi + v0yj + v0zk.  At time t, i.e. after a time interval ∆t = t, its velocity has changed by an amount ∆v = a∆t = at.  We can rewrite this in terms of the components as

∆vxi + ∆vyj + ∆vzk = axt i + ayt j + azt k.

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

∆vx = ax∆t,  ∆vy = ay∆t,   ∆vz = az∆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

vx = v0x + ∆vx = v0x + ax∆t,
vy = v0y + ay∆t,
vz = v0z∆t + az∆t,
or   v = v0 + a∆t.

Note: If the directions of v0 and a are different, the direction of v is different from the direction of v0.

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 = x0 + v0x∆t + ½ax∆t2,
y = y0 + v0y∆t + ½ay∆t2,
z = z0 + v0z∆t + ½az∆t2,
or    r = r0 + v0∆t + ½a∆t2.

Note: The directions of r0, v0, 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.

#### Problem:

At t = 0, a particle moving in the xy-plane with constant acceleration has a velocity v0 = (3i - 2j) m/s at the origin.  At t = 3 s, the particle's velocity is v = (9i + 7j) 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) - v0)/(3 s) = (9 m/s - 3 m/s)i/(3 s) + (7 m/s + 2 m/s)j/(3 s).
a
= (2 m/s2)i + (3 m/s2)j.

(b)  At t = 0 the particle is at the origin, r0 = 0.  Therefore r = v0t + ½at2 is its position at time t.
r = [(3 m/s)t + (1 m/s2)t2]i + [(2 m/s)t + (1.5 m/s2)t2]j.
x(t) = (3 m/s)t + (1 m/s2)t2,   y(t) = (2 m/s)t + (1.5 m/s2)t2.

#### Problem:

A particle originally located at the origin has an acceleration of a =  3j m/s2 and an initial velocity of v0 = 5i 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 v0 is given.  Its position at any time t is r = v0t + ½at2.
r = (5 m/s)ti + (1.5 m/s2)t2j.
The particles velocity at time t is v = v0 + at.
v = (5 m/s)i + (3 m/s2)tj.

(b)  At t = 2 s the coordinates of the particle are x = 10 m and y = 6 m.  Its speed is
v = (vx2 + vy2)½ = (25 + 36)½ m/s = 7.81 m/s.