In this section we will discuss the velocity and acceleration of an object inone-dimensional motion. Instantaneous velocityis the rate of change of the position of the object with respect to time. Therefore we can identify instantaneous velocity with a derivative. Instantaneous acceleration isthe rate of change of velocity with respect to time and can be computed as the derivative of the velocity function.
We will consider the simplest kind of motion that involves an object which moves along a line.
The object can be a car travelling along a straight stretch of a highway, a ball falling down from the Tower of Pisa, or a weight that is attached to a vibrating spring. We model the object as a point on the number line. We place the number line so that the positive direction coincides with the direction of motion that is selected as the positive direction. Letf(t)denote the position of the object at timet. Time is measured in units such as hours or seconds, and distance is measured in units such as miles or centimeters (we may not bother to indicate specic units in every example or problem). We will refer to this kind of mathematical model asone- dimensional motion, and to f asthe position function of the object in one dimensional motion.
132 CHAPTER 2. THE DERIVATIVE
ft 0
Figure 1: One-dimensional motion Let’s look at some examples.
Example 1 Letf(t) = 60tbe the position at timetof a car travelling along a straight stretch of a highway. Distances are measured in miles.
We have f(0) = 0, so that the origin of the number line corresponds to the position of the car at the time we start to monitor its motion. Figure 2 illustrates the motion of the car in space-time, i.e., the graph off in thety-plane. The graph off is the part of the liney= 60t that corresponds tot0. ¤
1 2 3 4 t
60 120 180 240 y
Figure 2: The motion of the car as illustrated in space-time
Example 2 Assume that a ball is dropped from a high tower (perhaps by Galileo from the top of the leaning tower of Pisa).
Let’s model the ball as a point on the y-axis that points downward and let the origin coincide with the point at which the ball is released, as illustrated in Figure 3.
yft 0
Figure 3
Assume that f(t) = 4.9t2 is the position of the ball (in meters), ift is such that the ball has not hit the ground yet. Figure 4 illustrates the motion in space-time. The graph ofy=tin the ty-plane is part of the parabolay= 4.9t2. ¤
1 2 3 t 20
40 y
y4.9t2
Figure 4: The motion of a falling object as illustrated in space-time
Example 3 Assume that a projectile is launched vertically from the ground level, rises up to a certain height, and falls back to the ground.
We assume that the motion is along a vertical line during the relevant time interval. Let y=f(t) = 196t4.9t2
be the position of the projectile above the ground at time t. We measure distances in meters and time in seconds. The motion is along they-axis. The positive direction coincides with the upward movement of the projectile.
y
yt
Figure 5: A projectile that is climbing vertically We have
f(t) = 0196t4.9t2=t= 0ort= 196 4.9 = 40.
Thus, the projectile is red from the ground level att= 0, and hits the ground att=T = 40 (seconds).
We can determine the maximum height of the projectile by completing the square: Since f(t) =4.9t2+ 196t=4.9¡
t240t¢
=4.9 (t20)2+ 1960,
the graph of f is part of a parabola with vertex at(20,1960). Thus, the rocket reaches a height of 1960 meters, then falls back, and hits the ground 40 seconds after it has been launched.
Figure 6 shows the graph off. ¤
20 40t
1000 1960 y
Figure 6: The path of a projectile in space-time
134 CHAPTER 2. THE DERIVATIVE
How should we determine the velocity of an object in one-dimensional motion? If the position function is linear as in Example 1, this is straightforward: iff(t) = 60tandt >0 represents a time increment,the average velocity of the car in the time interval[t,t]is
change in position
elapsed time = f(t+ t)f(t)
t = 60 (t+ t)60t
t = 60t
t = 60
(miles per hour.) This quantity is independent oftandt. We can say that the velocity at any instanttis 60 miles per hour. Note that 60 is also the slope of the line that is the graph of the linear functionf. Similarly, iff(t) =mt+b, then
f(t+ t)f(t)
t = [m(t+ t) +b][mt+b]
t
= mt+mt+bmtb
t = mt
t =m
for anyt andt, so that the velocity of the object has the constant valuemat any instantt.
This number is also the slope of the graph off.
Let’s now assume that the position function f is nonlinear, as in Example 2 and Example 3.
Let’s consider a specic instant t (you can imagine that time is frozen at t) and let t be an arbitrary positive time increment. As in the linear case, we can determine the average velocity over the time interval[t, t+ t]as
change in position
elapsed time = f(t+ t)f(t)
t .
In general, average velocity depends ontandt. It seems reasonable to dene the instantaneous velocity at the instanttas the limit of the average velocity as the time increment approaches 0:
Denition 1 The instantaneous velocityv(t)at the instanttof an object in one-dimensional motion is the derivative of the position functionf att:
v(t)= d f
dt(t)= lim
w0
f(t+ t)f(t)
t .
We may refer tov(t)simply as the velocity at the instantt. Since we have identied the rate of change of a function at a point as its derivative at that point,v(t)is the rate of change of the position function at the instant t.
Note that we have not restricted t to be positive in the above denition. Indeed, if t <0 then
f(t+ t)f(t)
t = f(t)f(t+ t)
(t) ,
so that the dierence quotient may be interpreted as the average velocity of the object on the time interval[t+ t, t].
Graphically, the average velocity over the time interval[t, t+ t]is the slope of the secant line that passes through the points (t, f(t)) and (t+ t, f(t+ t)) on the graph of the position function, and the instantaneous velocity at the instantt is the slope of the tangent line to the graph off at(t, f(t)).
t y
t, ft
t t t
t
ft tft f
Figure 7
Example 4 Letf(t) = 4.9t2, as in Example 2.
The average velocity of the ball over the time interval determined bytandt+ t is f(t+ t)f(t)
t = 4.9 (t+ t)24.9t2
t = 4.9
Ã
t2+ 2tt+ (t)2t2 t
!
= 4.9 (2t+ t). The instantaneous velocity of the ball at the instanttis the limit of the average velocity as the time incrementtapproaches 0:
v(t) = df
dt(t) = lim
t0
f(t+ t)f(t)
t = lim
t04.9 (2t+ t) = 9.8t
(meters per second).. Thus, velocity is not constant over time, unlike the case of a linear position function. Figure 8 displays the graph of the velocity function. ¤
1 2 3 4 t
20 40 v
vt9.8t
Figure 8: The velocity function of Example 4
If an object is moving in the direction that has been assigned as the positive direction, and t >0then
f(t+ t)f(t)
t 0.
Therefore,
v(t) = lim
t0
f(t+ t)f(t)
t 0
Similarly,v(t) 0 if the object is moving in the opposite direction. The speed of the object at timet is dened as the magnitude of velocity at timet, i.e.,|v(t)|.
Example 5 Let
f(t) = 196t4.9t2,
as in Example 3. Determine the velocity and the speed of the projectile as functions of t.
Interpret the sign of the velocity function with reference to the direction of motion of the projectile.
136 CHAPTER 2. THE DERIVATIVE Solution
a) As we discussed in Example 3, the relevant time interval is[0,40]. The velocity at the instant tis
v(t) = d dt
¡196t4.9t2¢= 1964.9 (2t) = 1969.8t.
Therefore,
v(t) = 0t= 196 9.8 = 20 (seconds). We have
v(t)>0if0< t <20andv(t)<0if 0< t <40.
Since the positive direction is upward in this example, the projectile climbs up in the time interval[0,20]and falls back towards the ground in the time interval [20,40]. The maximum height of the projectile is
f(20) = 196t4.9t2¯¯t=20= 1960meters.
The instantaneous velocity att= 20is 0. You can imagine that the projectile is momentarily stationary at that very instant, before it starts to fall back to earth. The velocity at the time of impact isv(40) =196meters/second. At that instant, speed is|196|= 196meters/second.
Figure 9 shows the graph of the velocity function. ¤
20 40t
196 196 v
v
Figure 9: The velocity function of Example 5
Intuitively, acceleration is the rate of change of velocity. We translate “rate of change” to
“derivative”:
Denition 2 Let v be the velocity function of an object in one-dimensional motion. The (instantaneous) acceleration a(t) of the object at the instant t is the derivative of the velocity att:
a(t) = d dtv(t).
Thus,acceleration is the second derivative of the position functionf(t):
a(t) = d
dtv(t) = d dt
d dtf(t)
ả
= d2 dt2f(t).
We may refer to a(t) simply as the acceleration at the instant t. Since the unit of veloc- ity is (unit of distance)/(unit of time), the unit of acceleration is (unit of distance)/(unit if time)/(unit of time)=(unit of distance)/(unit of time)2. For example, if distance is measured in meters and time is measured in seconds, acceleration is measured in meters/second/second
=meters/second2.
Example 6 Letf(t) = 60t, as in Example 1. We havev(t) = 60(miles/hour). The acceleration of the car is
a(t) = d
dtv(t) = d
dt(60) = 0.
Indeed, the velocity is a constant, so that its rate of change of is 0. ¤
Example 7 Letf(t) = 4.9t2, as in Example 4. We have v(t) = 9.8t(meters/second). There- fore, the acceleration of the falling ball is
a(t) = d
dtv(t) = d
dt(9.8t) = 9.8(meters/second/second).
The above expression for acceleration is consistent with the assumptions that the only force acting on the object is due to gravitational acceleration of 9.8 meters/second2, and that the opposing force due to air resistance can be neglected . Indeed, byNewton’s second law of motion,
Force = mass×acceleration.
Therefore, if the object has massm(kilograms), the force that is acting on the object due to gravitational accelerationg is the weightmgof the object. This force is 9.8m, if it is assumed thatg= 9.8meters/second2. Thus, we have the equation
ma(t) =mga(t) = 9.8
In Chapter 5 we will see that this expression for acceleration leads tov(t) = 9.8t and f(t) = 4.9t2.¤
Example 8 Let f(t) = 196t4.9t2, as in Example 5. We calculated the velocity as v(t) = 1969.8tTherefore,
a(t) =dv dt = d
dt(1969.8t) =9.8 The sign is()since the positive direction is upward in this example. ¤
Example 9 Assume thatf(t) = 2 cos (t)is the position of an object at timet. Thus, the object oscillates about the origin. Determine the velocity and acceleration functions.
Solution
The velocity at timetis
v(t) = d
dtf(t) = d
dt(2 cos (t)) =2 sin (t). The acceleration at timetis
a(t) = d
dtv(t) = d
dt(2 sin (t)) =2 cos (t). Note that
a(t) =f(t).
The motion is periodic with period2. Figure 10 displays the graphs of position, velocity and acceleration functions on the interval[0,2]. ¤
138 CHAPTER 2. THE DERIVATIVE
Π
2 Π 3Π2 2Πt
1 1
position
Π
2 Π 3Π2 2Πt
1 1
velocity
Π
2 Π 3Π2 2Πt
1 1
acceleration
Figure 10
Problems
In problems 1 - 4,f(t)is the position at timetof an object in one-dimensional motion.
a) Determinev(t), the velocity of the object at timet, anda(t), the acceleration of the object at timet.
b) Calculatev(t0)anda(t0).
1.
f(t) = 200t5t2, t0= 1 2.
f(t) = 5t2+ 100; t0= 4
3.
f(t) = 10 sin (t), t0=/6 4.
f(t) = 3 sin (t) + 8 cos (t), t0=/2