ONE OF THE most important applications of calculus is to motion in a straight line, which is called rectilinear motion.
Consider a particle moving in a straight line from a fixed point O to a given point P, and let t be the time elapsed. Then to each value of t there will correspond a distance s, which will be a function of t:
s = s(t).
When we know s(t), we have what is called the equation of motion.
Now, if the particle moves with constant velocity -- which is called uniform motion -- then we don't need calculus. In other words, if the equation of motion is
s = 22 t,
then at every instant of time, the velocity is 22 m/sec. For, the slope of that line, which is 22, is rate of change of s with respect to t, which by definition is the velocity.
In each 1 second of time, the particle moves a distance of 22 meters.
That is not a realistic picture, of course, because at 0 seconds the velocity is surely not 22 meters/sec. There must have been an acceleration
to that constant velocity. During that acceleration, the velocity was not constant. The graph was not a straight line.
The definition of instantaneous velocity
For any equation of motion s(t), we define what we call the instantaneous velocity at time t -- v(t) -- to be the limit of the average velocity, , between t and t + Δt, as Δt approaches 0.
The instantaneous velocity is the value of the slope of the tangent line at t.
Example 1. Let the following be the equation of motion:
s(t) = 6t 2 + t + 8.
Let t be measured in minutes and s in meters. And let 10:05 AM correspond to t = 0.
a) What is the instantaneous velocity at 10:05 AM?
b) What is the instantaneous velocity at 10:15 AM?
Answer. At 10:15, t = 10.
Instantaneous velocity is very different from ordinary velocity, which, to calculate, requires an interval of time. "Instantaneous velocity," like any limit, is defined at a specific value of time t. It is purely logical; it can never be observed or measured. To measure a velocity, it is necessary to know both a distance Δs and a time Δt, however small.
A body in motion is in motion during every interval of time in which it moves. If we say at every instant of time -- and assume that time is composed of instants -- then we invite the arrow paradox of Zeno. At any one instant, he argued, there is no change of time, no change of position, which means: no motion.
The definition of instantaneous velocity does not imply that time is composed of instants. It defines how to evaluate that velocity at any
See the Appendix, Is a line really composed of points.
Problem 1. It has been found by experiment that a body falling from rest under the influence of gravity, follows approximately this equation of motion:
s(t) = 4.9 t 2.
s is the distance fallen measured in meters; t is the time elapsed measured in seconds.
a) At the end of 3 seconds, how far has the body fallen?
To see the answer, pass your mouse over the colored area.
s(3) = 4.9 × 32 = 4.9 × 9 = 44.1 m.
b) What is its instantaneous velocity at the end of 3 seconds?
The second derivative
The derivative of y = f(x) -- y'(x) -- will itself be a function of x. That new function may itself be differentiable. If that is the case, we call the derivative of the first derivative of y the second derivative. The notation
Consider this equation of motion,
s(t) = 3t 2.
Then the first derivative is the velocity v:
The second derivative is the rate of change of the velocity with respect to time. That is called the acceleration a:
If t is measured in seconds and s in meters, then the units of velocity are meters per second, which we abbreviate as m/sec. The units of acceleration are then meters per second per second, which we abbreviate as m/sec2.
Problem 2. A body moves in a straight line according to this equation of motion:
s(t) = 10t 2 − 4t + 8,
where t is measured in seconds and s in meters.
a) What is its position at the end of 5 sec?
s(5) = 10· 52 − 4· 5 + 8 = 238 m.
b) What is the equation for its velocity v at any time t ?
c) What is its velocity v at the end of 5 seconds?
v(5) = 20· 5 − 4 = 96 m/sec.
d) What is the equation for its acceleration a at any time t ?
e) What is its acceleration at the end of 5 seconds?
a(5) = 20 m/sec2.
The acceleration is constant.
Problem 3. Under the influence of gravity, a body moves according to this equation of motion:
s(t) = ½gt 2 + s0
a) What is the physical significance of the constant s0?
It is the body's initial position, s(0).
b) How fast is the body moving after 5 seconds?
c) What is the physical significance of the constant g?
approximately 9.8 m/sec2.
a) If the radius of a circle is expanding, write the equation that shows
b) If the radius is expanding at the rate of 2 cm/min, how fast is the
when r = 15 cm:
Example 3. A boy is walking at the rate of 5 miles per hour toward the foot of a flag pole 60 feet high. At what rate is his distance from the top of the pole changing when he is 80 feet from its foot?
Solution. Draw a picture In all problems of this type, draw a picture.
Let the boy be at the point A, which is a distance x from the foot of the flag pole. Let s be his distance from the top of the pole.
The figure is a right triangle. Therefore,
Differentiate implicitly with respect to t.
write the minus sign because x is decreasing as he approaches the flag pole.
According to line 1),
Problem 4. The side of a square is a cm long, and is increasing at the rate of b cm per hour. How fast is the area increasing?
Problem 5. The side of an equilateral triangle is a cm long, and is increasing at the rate of b cm per hour. How fast is the area increasing?
a) The surface area S of a sphere is given by the formula S = 4πr2.
If r changes with time, how does S change?
If r changes with time, how does V change?
Problem 7. The base and height of a rectangle are b and h, and they are changing at the rates p, q respectively. Prove that the area A is changing at the rate bq + hp.
A = bh.
According to the product rule:
a) A ladder 50 feet long is leaning against a wall. If the foot of the
b) If the foot is being pulled away at the rate of 3 ft/min:
1) how fast is the top descending when the foot is 14 feet from the
2) when is the top descending at the rate of 4 ft/min?
When the bottom is 40 feet from the wall.
c) When will the top and bottom move at the same rate?
When the bottom is 25 feet from the wall.
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