166
Example
Question: Two objects have positions as functions of time given
by the equations
r
1
=
3t
2
x
x
+
+ty
y
and
r
2
=
3t
4
x
x
+
+ty
y
.
Find both objects’ accelerations using calculus. Could either
answer have been found without calculus.
Solution: Taking the first derivative of each component, we find
v
1
=
6tx
x
+
+y
y
v
2
=
12t
3
x
x
+
+y
y
,
and taking the derivatives again gives acceleration,
a
1
=
6x
x
a
2
=
36t
2
x
x
.
The first object’s acceleration could have been found without
calculus, simply by comparing the x and y coordinates with the
constant-acceleration equation
.
x=v
o
.
t+
1
2
a
.
t
2
. The second
equation, however, isn’t just a second-order polynomial in t, so
the acceleration isn’t constant, and we really did need calculus to
find the corresponding acceleration.
Example: a fire-extinguisher stunt on ice
Question: Prof. Puerile smuggles a fire extinguisher into a
skating rink. Climbing out onto the ice without any skates on, he
sits down and pushes off from the wall with his feet, acquiring an
initial velocity v
o
y
. At t=0, he then discharges the fire extin-
guisher at a 45-degree angle so that it applies a force to him that
is backward and to the left, i.e. along the negative y axis and the
positive x axis. The fire extinguisher’s force is strong at first, but
then dies down according to the equation |F|=b–ct, where b and
c are constants. Find the professor’s velocity as a function of
time.
Solution: Measured counterclockwise from the x axis, the angle
of the force vector becomes 315
°
. Breaking the force down into x
and y components, we have
F
x
=|F| cos 315
°
=
1
2
(b–ct)
F
y
=|F| sin 315
°
=
1
2
(–b+ct) .
In unit vector notation, this is
F=
1
2
(b–ct)
x
+
1
2
(–b+ct)
y
.
Newton’s second law gives
a=F/m
=
b–ct
2
m
x
+
–b+ct
2
m
y
.
Chapter 8Vectors and Motion
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