Arkansas Tech University
MATH 2934: Calculus III
Dr. Marcel B Finan
10.9
Motion in Space: Velocity and Acceleration
In this section we discuss how to find the velocity and acceleration vectors
from the parametric equations of the motion of the object.
Let ~r(t) be the position vector at time t and ~r(t + ∆t) the position at time
t + ∆t. Then the displacement vector between these two positions is ∆~r(t) =
~r(t + ∆t) − ~r(t) as shown in Figure 10.9.1. The average velocity over the
interval [t, t + ∆t] is defined by
Average velocity =
∆~
r(t)
∆t
and the instantaneous velocity is
∆~r(t)
~r(t + ∆t) − ~r(t)
~v (t) = r~0 (t) = lim
= lim
∆t→0 ∆t
∆t→0
∆t
whenever the limit exists. Note that ~v (t) is tangent to the object’s path of
motion.
Figure 10.9.1
We define the speed of the object to be the length of the vector ~v (t).
In a similar way, we define the acceleration vector to be
~v (t + ∆t) − ~v (t)
∆~v (t)
= lim
∆t→0
∆t→0 ∆t
∆t
~a(t) = lim
whenever the limit exists. We write ~a(t) = ~v 0 (t) = ~r 00 (t)
Example 10.9.1
Find the velocity, speed and acceleration of the motion with position function
~r(t) =< 3 cos (t2 ), 3 sin (t2 ), t2 > .
Solution.
The velocity vector is given by
~v (t) = x0 (t)~i + y 0 (t)~j + z 0 (t)~k = −6t sin (t2 )~i + 6t cos (t2 )~j + 2t~k.
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The speed is
||~v (t)|| =
√
p
√
(−6t sin t2 )2 + (6t cos t2 )2 + 4t2 = 40t2 = 2t 10.
The acceleration is the vector
~a(t) =x00 (t)~i + y 00 (t)~j + z 00 (t)~k
= − 6(sin (t2 ) + 2t2 cos (t2 ))~i + 6(cos (t2 ) − 2t2 sin (t2 ))~j + 2~k
Vector integrals can be used to find position vectors when velocity or acceleration vectors are known, as in the next example.
Example 10.9.2
Given the acceleration ~a(t) = ~i + t~kwith initial velocity ~v (0) = 5~j and initial
position ~r(0) = ~i, find the velocity and position vector functions.
Solution.
Since ~a(t) = v~0 (t), we have
Z
t2
~
(~i + t~k)dt = t~i + ~k + C.
2
Z
t2
(~i + t~k)dt = t~i + 5~j + ~k.
2
~v (t) =
~ Hence,
But ~v (0) = 5~j = C.
~v (t) =
Now, since ~v (t) = r~0 (t), we have
Z
t2
t2
t3
~
~r(t) = (t~i + 5~j + ~k)dt = ~i + 5t~j + + D.
2
2
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~ = ~i. Hence,
But ~r(0) = D
~r(t) = (
t2
t3
+ 1)~i + 5t~j +
2
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Uniform Circular Motion
A circular motion is the motion of an object along a path of radius R as
2
shown in Figure 10.9.2.
Figure 10.9.2
There are two ways to describe the motion of the object−linear and angular
speed. The linear speed v of the object is the rate at which the distance
traveled is changing. It is defined by the formula
v=
s
t
where s is the distance traveled on the circle.
The angular speed ω is the rate at which the central angle is changing. It is
given by
θ
ω= .
t
Its units are radians per seconds.
Thus, a parameterization of the motion is given by
~r(t) = R cos θ~i + R sin θ~j = R cos (ωt)~i + R sin (ωt)~j
where t is the time in seconds.
Now, since ||~r(t)|| = R, we have ~r(t) · r~0 (t) = 0 so that the linear velocity of
the object is always tangent to the circle.
The velocity is
~v (t) = −Rω sin (ωt)~i + Rω cos (ωt)~j.
The speed of the object is
p
||~v (t)|| = (Rω sin (ωt))2 + (Rω cos (ωt))2 = Rω.
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Hence, the speed of the object is constant and thus the name uniform.
Next, the acceleration vector is given by
~a(t) = ~v 0 (t) = −ω 2 (R cos (ωt)~i + R sin (ωt)~j) = −ω 2~r(t).
Since the acceleration is a multiple of the position, these vectors are parallel. Note the minus sign in the above expression means that the acceleration
points in the opposite direction of ~r(t) that is towards the center of the circle.
Since ~v is perpendicular to ~r, the acceleration is also perpendicular to ~v .
Force
Force is yet another example of a vector quantity. In order to move an object,
something must push on it. The strength of the “push” is the magnitude of
the force applied. However the direction of the push is important. If you wish
to move a chair in a room, it’s much more effective to push on it sideways
then to push down on it! In both cases, the same magnitude of force may
be applied to the chair, but the results will be substantially different. Hence
force has direction as well as magnitude and is therefore a vector.
The vector version of Newton’s second law of motion states that if, at
any time t, a force F~ (t) acts on an object of mass m producing an acceleration
~a(t) then
F~ (t) = m~a(t).
In the case of uniform circular motion, we have F~ (t) = −mω 2~r(t) which
shows that the force acts in the direction opposite to the radius vector and
therefore points toward the origin.
Example 10.9.3
A projectile of mass m = 50kg is fired with an initial speed of 500 m/s and
angle of elevation 30 degrees. Assuming air resistance is negligible (so the
only external force is due to gravity), find the following:
(a) the range of the projectile,
(b) the maximum height reached,
(c) the speed at impact.
Solution.
We set up the axes so that the projectile starts at the √
origin. The initial
◦~
◦~
velocity vector is ~v (0) = 500 cos 30 i + 500 sin 30 j = 250 3~i + 250~j.
The only force acting on the projective is the force of gravity that is pointing
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downward so that F~ (t) = m~a(t) = −mg~j, where g = 9.8 m/s2 . Hence,
~a(t) = −g~j. It follows that
~
~v (t) = −gt~j + C
√
~ = ~v (0) = 250 3~i + 250~j. Hence,
and C
√
~v (t) = 250 3~i + (250 − 9.8t)~j.
The vector position of the projectile is
Z
√
√
~
~r(t) = (250 3~i + (250 − 9.8t)~j)dt = 250 3t~i + (250t − 4.9t2 )~j + D.
~ = ~r(0) = ~0. Hence,
Since the projectile starts at the origin, we have D
√
~r(t) = 250 3t~i + (250t − 4.9t2 )~j.
2
√
x√
x
2
√
Letting x = 250 3t and y = 250t − 4.9t we find y = 3 − 4.9 250 3 =
√x
3
4.9
− 187,500
x2 which is a parabola in the xy−plane.
(a) The range of the projectile is the horizontal distance it will travel. To
calculate this, we need to know when it hits the ground. This occurs when
250t − 4.9t2 = 0. Solving this equation√we find t = 0 and t = 51 seconds.
Hence, the range of the projectile is 250 3(51) = 22, 083 meters or about 22
km.
(b) The maximum height is the maximum of the function y = 250t − 4.9t2
which occurs at the vertex of this parabola, i.e., at t = 125
≈ 25.5 and
4.9
y ≈ 250(25.5) − 4.9(25.5)2 ≈ 3, 189m.
q
√
(c) The speed at the impact is ||~v (51)|| = (250 3)2 + (250 − 9.81(51))2 ≈
500 m/s
Tangential and normal components of acceleration
When studying the motion of an object, it is often useful to resolve the acceleration vector into two components−one in the direction of the tangent
(the direction we are traveling) and the normal (the direction the curve is
turning). If v(t) = ||~v (t)|| is the speed of an object whose position is given
by the vector function ~r(t), then
r~0 (t)
~v (t)
T~ (t) =
=
0
v(t)
||r~ (t)||
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or
~v (t) = T~ (t)v(t).
Differentiating this equation with respect to t, we find
~a(t) = v 0 (t)T~ (t) + v(t)T~ 0 (t).
Using the expression of the curvature
κ=
||T~ 0 (t)||
||r~0 (t)||
we find ||T~ 0 (t)|| = κv(t). On the other hand, we have
~0
~0
~ (t).
~ (t) = T (t) = T (t) =⇒ T~ 0 (t) = κv(t)N
N
0
~
κv(t)
||T (t)||
Hence, ~a(t) can be resolved in terms of the normal and tangent vectors T~ (t)
~ (t) as
and N
~ (t).
~a(t) = v 0 (t)T~ (t) + κv 2 (t)N
Letting, aT = v 0 (t) be the tangential component and aN = κv 2 (t) be the
normal component, the resolution of ~a(t) is described geometrically in Figure
10.9.3.
Figure 10.9.3
Next, we express aT = v 0 (t) and aN = κv 2 t in terms of ~r(t), r~0 (t), and r~00 (t).
We have
~ (t))
~v · ~a(t) =v(t)T~ (t) · (v 0 T~ (t) + κv 2 (t)N
~
=v(t)v 0 (t)T~ · T~ + κv 3 (t)T~ · N
=v(t)v 0 (t)||T~ ||2 = v(t)v 0 (t).
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~0
~00
·~a(t)
r (t)
Hence, aT = v 0 (t) = ~vv(t)
= r ||(t)·
.
r~0 (t)||
On the other hand, using the formula
κ=
||r~0 (t) × r~00 (t)||
||r~0 (t)||3
we arrive at
aN = κv 2 (t) =
||r~0 (t) × r~00 (t)||
.
||r~0 (t)||
Example 10.9.4
Find the tangential and normal components of the acceleration vector of
~r(t) = (3t − t3 )~i + 3t2~j.
Solution.
Straight forward calculation, we find
r~0 (t) =(3 − 3t2 )~i + 6t~j
p
√
||r~0 (t)|| = (3 − 3t2 )2 + 36t2 = 9 + 18t2 + 9t4
p
= 9(1 + t2 )2 = 3(1 + t2 )
r~00 (t) = − 6t~i + 6~j
r~0 (t) × r~00 (t) =(18t2 + 18)~k
r~0 (t) · r~00 (t) =18t3 + 18t
r~0 (t) · r~00 (t)
18t3 + 18t
aT =
=
= 6t
3 + 3t2
||r~0 (t)||
||r~0 (t) × r~00 (t)||
18t2 + 18
aN =
=
=6
3(1 + t2 )
||r~0 (t)||
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