Suppose that the position vector function for a particle is given by
= 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.
(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)1/2 = (1 + 0.25)1/2 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.
(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)|
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 = axi + ayj + azk
= (ax, ay, az) = constant. Since
constant, the components ax, ay, and az are
constant and the average acceleration is equal to the instantaneous
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
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.
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.
(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 + (1/2)at2 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.
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.
(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 = (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)1/2 = (25 + 36)1/2 m/s = 7.81 m/s.