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