## The Calculus Primer (2011)

### Part IV. Using the Derivative

### Chapter 14. DISTANCE, VELOCITY, AND ACCELERATION

**4—4. Distance and Speed.** The distance traveled by a moving object is its *displacement* If this displacement is achieved by motion in a straight line from its original position to a final position, it is referred to as *rectilinear motion*. The *speed* with which an object moves is the time rate of change—how far in a given unit of time. If in addition to the numerical value of the time rate of change the *direction* of motion is also specified, the rate is called the *velocity*. Thus, a plane flies with a speed of 150 miles per hour, but with a velocity of 150 miles per hour northeast.

If the distance is designated by *s*, the speed by *v*, and the time by *t*, then clearly, from our definitions,

or speed (velocity) equals the derivative of the distance with respect to the time. It is understood, of course, that the motion of a body may be *uniform* or *non-uniform*. In general, *s* is a function of *t*, or *s* = *f*(*t*).

If the motion is uniform, = *k*, or *s* = *kt*; and = *k*. This is the familiar *R* = formula of elementary algebra, where *R* equals the rate, or .

EXAMPLE 1.An object moves according to the formula *s* = 30*t* + 5*t*^{2}. Find (a) the distance traveled in 4 sec; (b) the distance traveled in the fourth second; (c) its velocity at the end of the fourth second.

*Solution*.

(a)When *t* = 4, *s* = (30)(4) + 5(4)^{2} = 200 ft.

(b)When *t* = 3, *s* = (30)(3) + 5(3)^{2} = 135 ft.

Hence, during fourth second, the distance covered equals 200 − 135 = 65 ft.

(c)Velocity = = 30 + 10*t*;

when *t* = 4, = 30 + 10(4) = 70 ft./sec.

EXAMPLE 2.A missile thrown vertically upward is moving in such a way that the height after *t* seconds is given by *h* = 192*t* − 16*t*^{2}. Find (a) the velocity at the end of 5 sec; (b) the velocity at the end of 8 sec; (c) the velocity at the end of 6 sec; (d) what is the maximum height reached?

*Solution*.

*h* = 192*t* − 16*t*^{2}.

= 192 − 35*t*.

(a)When *t* = 5, = 192 − (32)(5) = 32 ft./sec.

(b)When *t* = 8, = 192 − (32) (8) = −64 ft./sec;

the fact that the velocity is negative means that the missile is now moving *downward*, or falling.

(c)When *t* = 6, = 192 − (32)(6) = 0;

this means that the missile is moving neither upward nor downward.

(d)The maximum height is attained when = 0, or at *t* = 6; at *t* = 6, *h* = (192)(6) − 16(6)^{2} = 576 ft.

**4—5. Velocity and Acceleration.** In motion that is not uniform, the velocity is either increasing or decreasing. The rate at which the velocity is changing is called *acceleration;* when the velocity is decreasing, that is, when the acceleration is negative, it is frequently called *deceleration*.

Since acceleration, usually denoted by *a*, is the rate of change of velocity, we may write

but since velocity is the rate of change of distance, or

we have

The symbol is called the *second derivative* of the function *s* = *f*(*t*). It should not be confused with which is the square of the derivative. A second derivative is thus *the derivative of a derivative*. This is to be expected, since it is the rate at which another rate is changing; or, the rate of change of a rate of change.

EXAMPLE.A body is moving according to the formula *s* = *t*^{3} + 4*t*^{2}. Find the distance traveled, the velocity, and the acceleration at the instant when *t* = 3.

*Solution*.

(a)*s* = *t*^{3} + 4*t*^{2},

*s*_{3}= (3)^{3} + 4(3)^{2} = 63ft.

(b)*v* = = 3*t*^{2} + 8*t*,

*v*_{3} = 3(3)^{2} + 8(3) = 51 ft./sec.

(c)*a* = = 6*t* + 8,

*a*_{3} = 6(3) + 8 = 26 ft./sec./sec.

**4—6. Laws of a Falling Body.** It can be shown from physics that a body when falling freely from rest near the earth’s surface, if air resistance is disregarded, follows the law

*s* = *gt*^{2},(1)

where *g* is the constant of “gravitational acceleration,” and equals about 32.2 feet per second per second when *s* is measured in feet and *t* is measured in seconds. From equation (1) we get:

or*v* = *gt*;(2)

and*a* = (*v*) = (*gt*) = *g*,

or*a* = *g*.(3)

Also, from (1):*s* = *gt*^{2} = *t*(*gt*);

but from (2):*gt* = *v*;

hence*s* = *tv*,or2*s* = *vt*;

multiplying this last equation by the equation *gt* = *v*:

2*sgt* = *v*^{2}*t*

or*v*^{2} = 2*gs*.(4)

Equations (1), (2), and (4) are often referred to as the *laws of a freely-falling body*.

EXAMPLE.If a body falls freely from rest, find (a) its velocity at the end of 5 sec.; (b) the distance fallen in 10 sec.; (c) its velocity after it has fallen 40 feet.

*Solution*.

(a)From (2), *v* = *gt* = (32.2)(5) = 161 ft./sec.

(b)From (1), *s* = *gt*^{2} = (32.2)(100) = 1610ft.

(c)From (4), *v*^{2} = 2*gs*,

or*v* = = 50.8ft./sec.

**4—7. Successive Derivatives.** Differentiation may be repeated more than once. Thus, , or *f*″ (*x*), is the derivative of the derivative of *y* = *f*(*x*); that is,

EXAMPLE 1.Find the fourth derivative of *y* = 2*x*^{5} + 3*x*^{4} − 10*x*^{2} + 5.

*Solution*. = 10*x*^{4} + 12*x*^{3} − 20*x*;

EXAMPLE 2.Find the third derivative of *y* =

*Solution*.*y* = 3*x*^{−5}.

EXAMPLE 3.Find the second derivative of *y* = .

*Solution*.

**EXERCISE 4—2**

**1.** Find the second derivative of each of the following:

(a) *y* = *x*^{5} + 3*x*^{3} − 4*x*

(b) *s* = 50*t*^{2} − 2*t*^{5}

**2.** Find the third derivative of

(b) *z* = *t*(10 − *t*^{2})

**3.** Find the fourth derivative of

(a) *y* = *x*^{6} + 20 +

(b) *s* = *t*^{5} − *t*^{2}

**4.** Given the following equations of rectilinear motion, find the distance, velocity, and acceleration at the instant indicated in each case:

(a) *s* = 3*t* + *t*^{3}, *t* = 2

(b) *s* = 10*t* − 5*t*^{2}, *t* = 1

(c) *s* = 2*t*^{3} − 6*t*^{2}, *t* = 1

**5.** An object falls from rest; using the laws of a falling body (§4—6), find (a) its velocity at the end of 10 sec.; (b) its velocity after it has fallen 20 ft.; (c) its acceleration at the instant when *t* = 6; and (d) the distance it has fallen at the end of 4 seconds. (Use *g* = 32.2.)

**6.** If an object starts with an initial velocity of *v*_{0} and moves in a straight line with any given constant acceleration *a* (positive or negative), the distance traveled during a time *t* elapsed since the beginning of its motion is given by the formula

*s* = *v*_{0}*t* + *at*^{2}.

Prove: (a) *v* = *v*_{0} + *at*, and (b) (*v*^{2} − *v*_{0}^{2}) = *as*.

**7.** Using the formulas from Problem 6, find the acceleration (assumed constant) with which a train, starting from rest, acquires a velocity of 60 miles per hour in 4 minutes. How far does it travel in that time?

**8.** An automobile moving at 40 miles an hour is brought to rest uniformly in 2 minutes. Find (a) its constant retardation (negative acceleration), and (b) how far it moves in that time.

**9.** Neglecting air resistance, the height (*h* ft.) reached in *t* seconds by an object projected vertically upward with an initial velocity of *v*_{0} ft. per sec. is equal to *h* = *v*_{0}*t* − 16.1*t*^{2}. Find (a) its velocity and acceleration at any instant; (b) its velocity and acceleration at the end of 3 seconds when the initial velocity equals 200 ft. per sec.; and (c) its velocity and acceleration at the end of 12 seconds.

**10.** A projectile is shot vertically upward with an initial velocity of 1200 ft./sec. Find (a) its velocity at the end of 15 sec.; (b) for how long after it is fired will it continue to rise? (Use formula from Problem 9.)