3.3
Velocity and Rate of Change
Suppose we drop a rock from a cliff in the High Sierras. There is an equation
that we can use to determine the position of a falling object after t seconds.
If s(t) is the position in meters of a falling object after t seconds then we
have the formula
s(t) = 4.9t2 m.
For example, after 1 second, the rock has fallen s(1) = 4.9(1)2 = 4.9 m.
After 2 seconds, the rock has fallen s(2) = 4.9(2)2 = 4.9(4) = 19.6 m.
In fact, this formula can be used to estimate the height of a cliff. We can
make the approximation that s(t) = 5t2 m. Then we can drop a rock off of
a cliff and record how long it takes for the rock to fall. Say the rock takes 5
seconds to fall. In that case the distance that the rock fell is approximately
s(10) = 5 ∗ (5)2 = 125 m. So the distance from the top of the cliff to the
base is about 125 meters.
Now let’s talk about velocity. Using this formula for position, or the distance that a falling object has traveled, we can calculate the average velocity
of an object.
Suppose we are interested in knowing how fast a falling object is going
after it has fallen a seconds. We have the following formula for average
velocity.
vavg =
s(t2 ) − s(t1 )
t2 − t1
So the units for average velocity will be meters per second in our example.
To find how fast the falling object is going after it has traveled a seconds, we
can let t1 = a and t2 = t. This gives
vavg =
s(t) − s(a)
t−a
Common sense tells us that to get the best approximation possible, we
should make the interval over which we find the average velocity as small
as possible. We would like to take the limit of the average velocity as t
1
approaches a. This is how we define instantaneous velocity. The instantaneous velocity of of a particle with position function s(t) is given by
v(t) = lim
t→a
s(t) − s(a)
.
t−a
This is the formula for derivative of the position function s(t). Therefore,
instantaneous velocity is the derivative of the position function. That is,
v(t) = s0 (t)
Example 1 Suppose that a ball is dropped from the upper deck of the CN
Tower, 450 meters above the ground. The position function of the ball is
given by s(t) = 4.9t2 meters after t seconds.
1. What is the velocity of the ball after 5 seconds?
2. How fast is the ball traveling when it hits the ground?
Solution We can find a formula for velocity at time t seconds.
v(t) = s0 (t) = 4.9(2t) = 9.8t m/s
1. After 5 seconds, we use the formula with t = 5, so the velocity is
v(5) = 9.8(5) = 49 m/s.
2. Because the observation deck is 450 m above the ground, the ball will
hit the ground at time t1 when s(t1 ) = 450. That is,
4.9t21 = 450
This gives
t21 =
450
and t1 ≈ 9.6s
4.9
The velocity of the ball as it hits the ground is therefore
s
v(t1 ) = 9.8t1 = 9.8
2
450
≈ 94 m/s
4.9
Example 2 The position function of a particle is given by the equation
s = f (t) = t3 − 6t2 + 9t
where t is measured in seconds and s in meters.
1. Find the velocity at time t.
2. What is the velocity after 2s? After 4s?
3. When is the particle at rest?
4. When is the particle moving forward (that is, in the positive direction)?
Solution
1. Velocity is the derivative of the position function, s(t). Therefore,
v(t) = s0 (t) = 3t2 − 12t + 9
2. We can use the formula for velocity to find the velocity at t = 2 and
t = 4 seconds.
v(2) = 3(2)2 − 12(2) + 9 = −3m/s
v(4) = 3(4)2 − 12(4) + 9 = 9m/s
3. The particle is at rest when v(t) = 0, that is,
3t2 − 12t + 9 = 0
3(t2 − 4t + 3) = 0
3(t − 1)(t − 3) = 0
t = 1, t = 3
So the particle is at rest at t = 1s and t = 3 seconds.
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4. When the velocity is positive, the particle moves forward; when the
velocity is negative, the particle moves backward. We can make a
chart to determine on which intervals the function v(t) is positive or
negative. We already know that v = 0 at t = 1 and t = 3.
v=0
v=0
v(t) = 3(t − 1)(t − 3)
pos
neg
pos
(t − 3)
neg
neg
pos
(t − 1)
neg
pos
pos
-
0
1
2
3
4
5
The velocity is positive, and therefore the particle moves forward, for
0 < t < 1 or 3 < t.
If the position of an object is given by the function s(t), then we said that
the velocity is the derivative of the position function. That is, v(t) = s0 (t).
Derivative is rate of change. The rate of change of velocity is acceleration.
We define acceleration as a(t) = v 0 (t) = s00 (t). The acceleration function is
the derivative of the velocity function, or the second derivative of position
function.
Example 3 The position function of a particle is given by the equation
s = f (t) = t3 − 6t2 + 9t. Find the velocity and acceleration functions for the
particle.
Solution
v(t) = s0 (t) = 3t2 − 12t + 9
a(t) = v 0 (t) = 6t − 12
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