Motion in 2 and 3 dimensions

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

Problems:

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 = 1m/s, b = 1m, c = 0.125m/s2, and d = 1m.
(a)  Calculate the average velocity during the time interval t = 2s to t = 4s.
(b)  Determine the velocity and speed at t = 2s.
Solution:
(a) The position of the particle is given to us as a function of time.  At t = 2s the position of the particle is
r(2s) = [(1m/s)2s + 1m]i  + [(0.125m/s2)4s2 + 1m]j = 3mi + 1.5mj.
At t = 4s its position is
r(4s)=[(1m/s)4s + 1m]i + [(0.125m/s2)16s2 + 1m]j = 5mi + 3mj.
The average velocity of the particle between 2 an 4 seconds is
v = Dr/Dt = (r(4s) - r(2s))/2s,
v = (5m - 3m)i/2s + (3m - 1.5m)j/2s = (1m/s)i + (0.75m/s)j.
(b) The instantaneous velocity of the particle is
v = dr/dt = [d(x(t)/dt]i + [dy(t)/dt]j = ai + 2ctj.
At t = 2s the instantaneous velocity is v(2s) = (1m/s)i + (0.5m/s)j.
The speed at 2 seconds is .
The coordinates of an object moving in the xy-plane vary with time according to the equations 
x = (-5m)sin(t) and y = (4m) - (5m)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 = (-5m/s)cos(t), vy = dy/dt = (5m/s)sin(t).  
At t = 0 we have v = (-5m/s)i.
Differentiating the velocity vector with respect to time we find 
ax = dvx/dt = (5m/s2)sin(t), ay = dvy/dt = (5m/s2)cos(t).  
At t = 0 we have a = (5m/s2)j.
(b) v = (-5m/s)cos(t)i + (5m/s)sin(t)j, a = (5m/s2)sin(t)i + (5m/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 = 4m.

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 be 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 Dt = t, its velocity has changed by an amount Dv = aDt = at.  We can rewrite this in terms of the components as

Dvxi + Dvyj + Dvzk = axti + aytj + aztk.


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

Dvx = axt,   Dvy = ayt,    Dvz = azt.

We therefore have at time t:

vx = v0x + Dvx = v0x + axt,    vy = v0y + ayt,   vz = v0zt + azt,    or

v = v0 + at.

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

Using the same arguments we made in module 2, we can now find the position of the particle at time t.

x - x0 = v0xt + (1/2)axt2,    y - y0 = v0yt + (1/2) ayt2,    z - z0 = v0zt + (1/2)azt2,

or

x = x0 +  v0xt + (1/2)axt2,    y = y0 + v0yt + (1/2) ayt2,    z = z0 + v0zt + (1/2)azt2,    or

r = r0 + v0t + (1/2)at2.

Note: The directions of r0, v0, a, and r may all be different.

Problems:

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 = 3s, 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 = 3s.
a = Dv/Dt = (v(3s) - v0)/3s = (9m/s - 3m/s)i/3s + (7m/s + 2m/s)j/3s.
a = (2m/s2)i + (3m/s2)j.
(b) At t = 0 the particle is at the origin, r0 = 0.  Therefore r = v0t + (1/2)at2 is its position at time t.
r = [(3m/s)t + (1m/s2)t2]i + [(2m/s)t + (1.5m/s2)t2]j.
x(t) = (3m/s)t + (1m/s2)t2,   y(t) = (2m/s)t + (1.5m/s2)t2.
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 = 2s.
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 + (1/2)at2.
r = (5m/s)ti + (1.5m/s2)t2j.
The particles velocity at time t is v = v0 + at.
v = (5m/s)i + (3m/s2)tj.
At t = 2s the coordinates of the particle are x = 10m and y = 6m.  Its speed is

Links to other Web Materials:

Motion in 2 Dimensions

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