# Velocity | MS-3 | Math 131, ApplicationsMotion: Velocity and Net Change

```math 131, applications
motion: velocity and net change
Motion: Velocity and Net Change
In Calculus I you interpreted the first and second derivatives as velocity and acceleration in the context of motion. So let’s apply the initial value problem results to
motion problems. Recall that
• s(t) = position at time t.
• s0 (t) = v(t) = velocity at time t.
• s00 (t) = v0 (t) = a(t) = acceleration at time t.
Therefore
R
•
a(t) dt = v(t) + c1 = velocity.
R
• v(t) dt = s(t) + c2 = position at time t.
We will need to use additional information to evaluate the constants c1 and c2 .
EXAMPLE 6.1. Suppose that the acceleration of an object is given by a(t) = 2 − cos t for
t ≥ 0 with
• v(0) = 1, this is also denoted v0
• s(0) = 3, this is also denoted s0 .
Find s(t).
SOLUTION. First find v(t) which is the antiderivative of a(t).
Z
Z
v(t) =
a(t) dt =
2 − cos t dt = 2t − sin t + c1 .
Now use the initial value for v(t) to solve for c1 :
v(0) = 0 − 0 + c1 = 1 ⇒ c1 = 1.
Therefore, v(t) = 2t − sin t + 1. Now solve for s(t) by taking the antiderivative of v(t).
Z
s(t) =
Z
v(t) dt =
2t − sin t + 1 dt = t2 + cos t + t + c2
Now use the initial value of s to solve for c2 :
s(0) = 0 + cos 0 + c2 = 3 ⇒ 1 + c2 = 3 ⇒ c2 = 2.
So s(t) = t2 + cos t + 2t + 2.
EXAMPLE 6.2. If acceleration is given by a(t) = 10 + 3t − 3t2 , find the exact position
function if s(0) = 1 and s(2) = 11.
SOLUTION. First
v(t) =
Z
Now
s(t) =
a(t) dt =
Z
Z
10 + 3t − 3t2 dt = 10t + 23 t2 − t3 + c.
10t + 23 t2 − t3 + c dt = 5t2 + 12 t3 − 14 t4 + ct + d.
But s(0) = 0 + 0 − 0 + 0 + d = 1 so d = 1. Then s(2) = 20 + 4 − 4 + 2c + 1 = 11 so
2c = −10 ⇒ c = −5. Thus, s(t) = 5t2 + 21 t3 − 14 t4 − 5t + 1.
EXAMPLE 6.3. If acceleration is given by a(t) = sin t + cos t, find the position function
if s(0) = 1 and s(2π ) = −1.
SOLUTION. First
v(t) =
Now
s(t) =
Z
Z
a(t) dt =
Z
sin t + cos t dt = − cos t + sin t + c.
− cos t + sin t + c dt = − sin t − cos t + ct + d.
But s(0) = 0 − 1 + 0 + 0 + d = 1 so d = 2. Then s(2π ) = 0 − 1 + 2cπ + 2 = −1 so
2πc = −2 ⇒ c = − π1 + 2. Thus, s(t) = cos t + sin t − π1 t.
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math 131, applications
motion: velocity and net change
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Displacement vs Distance Travelled
DEFINITION 6.1. The displacement of an object between times t = a and a later time t = b is
s(b) − s( a) =
Z b
a
v(t) dt.
The distance travelled by an object between times t = a and a later time t = b is
Distance travelled =
Z b
a
|v(t)| dt.
In other words, displacement is the net
area between the velocity curve and
the horizontal axis, while the distance
travelled is the total area between the
velocity curve and the horizontal axis.
EXAMPLE 6.4. Suppose an object moves with velocity 2t2 − 12t + 16 km/hr.
(1) Determine the displacement of the object on the time interval [1, 3] and [0, 4]
(2) Determine the distance travelled on [0, 4]
SOLUTION. (1) The displacement is easy to calculate: For the interval [0, 4], On [0, 4],
s (4) − s (0) =
Z 4
0
2t2 − 12t + 16 dt =
4
128
32
2t3
− 6t2 + 16t =
− 96 + 64 − 0 = .
3
3
3
0
On [1, 3],
s (3) − s (1) =
Z 3
1
2t2 − 12t + 16 dt =
4
2t3
2
4
− 6t2 + 16t = (18 − 54 + 48) −
− 6 + 16 = .
3
3
3
1
(2) The distance travelled is harder to determine since we need to integrate |v(t)|.
We must first determine where v(t) is positive and negative.
2t2 − 12t + 16 + + + 0 − − − 0 ++
2t2 − 12t + 16 = 2(t2 − 6t + 8) = 2(t − 2)(t − 4) = 0 ⇒ t = 2, 4.
2
4
The number line to the right shows that 2t2 − 12t + 16 ≤ 0 only [2, 4]. We can
now find the distance travelled (total area) by splitting the interval into two
pieces [0, 2] and [2, 4], changing the sign of v(t) on the second piece to obtain
the absolute value of v(t).
Dist Trav = Total Area =
Z 4
0
16
|2t2 − 12t + 16| dt
12
Z 2
2
Z 4
2
2t − 12t + 16 dt +
−(2t − 12t + 16) dt
2
3
3
4 2t
2t
2
=−
− 6t2 + 16t −
− 6t2 + 16t
3
3
0
2
40 8
+ = 16.
=
3
3
=
0
EXAMPLE 6.5. Suppose an object moves with velocity t3 − 5t2 + 4t m/s.
(1) Determine the displacement of the object on the time interval [0, 6] and interpret
8
4
0
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4
−4
Figure 6.1: The distance travelled on
[0, 4] is the area under the absolute
value of the velocity curve.
(2) Determine the distance travelled on [0, 6]
SOLUTION. (1) For displacement on [0, 6],
Z 6
0
6
5t3
t4
2
−
+ 2t = (324 − 360 + 72) − 0 = 36.
t − 5t + 4t dt =
4
3
0
3
2
(2) For the distance travelled we must first determine where v(t) is positive and
negative.
t3 − 5t2 + 4t = t(t2 − 5t + 4) = t(t − 1)(t − 4) = 0 ⇒ t = 0, 1, 4.
t3 − 5t2 + 4t 0 + 0 − − − 0 ++
0
1
4
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math 131, applications
motion: velocity and net change
3
The number line to the right shows that t3 − 5t2 + 4t ≤ 0 only [1, 4]. We can now
find the distance travelled (total area) by splitting the interval into three pieces
[0, 1], [1, 4] and [4, 6], changing the sign of v(t) on the second piece to obtain the
absolute value of v(t).
Dist Trav =
Z 6
0
|2t2 − 12t + 16| dt
Z 1
3
Z 4
2
3
2
Z 6
3
2
t − 5t + 4t dt −
t − 5t + 4t dt +
t − 5t + 4t dt
# "1
# 4"
#
"0
1
4
6
5t3
5t3
5t3
t4
t4
t4
2
2
2
−
+ 2t −
−
+ 2t +
−
+ 2t =
4
3
4
3
4
3
0
1
4
=
45 140
117
7
+
+
=
.
=
12
4
3
2
54
44
34
24
14
4
−6
Constant Acceleration: Gravity
In many motion problems the acceleration is constant. This happens when an
object is thrown or dropped and the only acceleration is due to gravity. In such a
situation we have
• a(t) = a, constant acceleration
• with initial velocity v(0) = v0
• and initial position s(0) = s0 .
Then
v(t) =
Z
a(t) dt =
Z
a dt = at + c.
But
v (0) = a · 0 + c = v0 ⇒ c = v0 .
So
v(t) = at + v0 .
Next,
s(t) =
Z
v(t) dt =
Z
at + v0 dt = 21 at2 + v0 t + c.
At time t = 0,
s(0) = 12 a(0)2 + v0 (0) + c = s0 ⇒ c = s0 .
Therefore
s(t) = 12 at2 + v0 t + s0 .
EXAMPLE 6.6. Suppose a ball is thrown with initial velocity 96 ft/s from a roof top 432
feet high. The acceleration due to gravity is constant a(t) = −32 ft/s2 . Find v(t) and
s(t). Then find the maximum height of the ball and the time when the ball hits the
ground.
SOLUTION. Recognizing that v0 = 96 and s0 = 432 and that the acceleration is
constant, we may use the general formulas we just developed.
v(t) = at + v0 = −32t + 96
and
s(t) = 21 at2 + v0 t + s0 = −16t2 + 96t + 432.
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math 131, applications
motion: velocity and net change
The max height occurs when the velocity is 0 (when the ball stops rising):
v(t) = −32t + 96 = 0 ⇒ t = 3 ⇒ s(3) = −144 + 288 + 432 = 576 ft.
The ball hits the ground when s(t) = 0.
s(t) = −16t2 + 96t + 432 = −16(t2 − 6t − 27) = −16(t − 9)(t + 3) = 0.
So t = 9 only (since t = −3 does not make sense).
EXAMPLE 6.7. A person drops a stone from a bridge. What is the height (in feet) of the
bridge if the person hears the splash 5 seconds after dropping it?
SOLUTION. Here’s what we know. v0 = 0 (dropped) and s(5) = 0 (hits water). And
we know acceleration is constant, a = −32 ft/s2 . We want to find the height of the
bridge, which is just s0 . Use our constant acceleration motion formulas to solve for a.
v(t) = at + v0 = −32t
and
s(t) = 21 at2 + v0 t + s0 = −16t2 + s0 .
Now we use the position we know: s(5) = 0.
s(5) = −16(5)2 + s0 ⇒ s0 = 400 ft.
Notice that we did not need to use the velocity function.
YOU TRY IT 6.1 (Extra Credit). In the previous problem we did not take into account that
sound does not travel instantaneously in your calculation above. Assume that sound travels at 1120 ft/s. What is the height (in feet) of the bridge if the person hears the splash 5
seconds after dropping it?
EXAMPLE 6.8. Here’s a variation. This time we will use metric units. Suppose a ball
is thrown with unknown initial velocity v0 m/s from a roof top 49 meters high and
the position of the ball at time t = 3 is s(3) = 0. The acceleration due to gravity is
constant a(t) = −9.8 m/s2 . Find v(t) and s(t).
SOLUTION. This time v0 is unknown but s0 = 49 and s(3) = 0. Again the acceleration
is constant so we may use the general formulas for this situation.
v(t) = at + v0 = −9.8t + v0
and
s(t) = 21 at2 + v0 t + s0 = −4.9t2 + v0 t + 49.
But we know that
s(3) = −4.9(3)2 + v0 · 3 + 49 = 0
which means
3v0 = 4.9(9) − 4.9(10) = −4.9 ⇒ v0 = −4.9/3.
So
v(t) = −9.8t −
49
30
and
s(t) = −4.9t2 −
49
30 t + 49.
Interpret v0 = −4.9/3.
EXAMPLE 6.9. Mo Green is attempting to run the 100m dash in the Geneva Invita-
tional Track Meet in 9.8 seconds. He wants to run in a way that his acceleration is
constant, a, over the entire race. Determine his velocity function. (a will still appear as
an unknown constant.) Determine his position function. There should be no unknown
constants in your equation at this point. What is his velocity at the end of the race?
Do you think this is realistic?
bridge be higher or lower than in the
preceding example? Why?
4
math 131, applications
motion: velocity and net change
SOLUTION. We have: constant acceleration = a m/s2 ; v0 = 0 m/s; s0 = 0 m. So
v(t) = at + v0 = at
and
s(t) = 21 at2 + v0 t + s0 = 12 at2 .
200
= 100, so a = (9.8
= 2.0825 m/s2 . So s(t) = 2.0825t2 . Mo’s
)2
velocity at the end of the race is v(9.8) = a · 9.8 = 2.0825(9.8) = 20.41 m/s. . . not
realistic.
But s(9.8) =
1
2
2 a (9.8)
EXAMPLE 6.10. A stone dropped off a cliff hits the ground with speed of 120 ft/s.
What was the height of the cliff?
SOLUTION. Notice that v0 = 0 (dropped!) and s0 is unknown but is equal to the cliff
height, and that the acceleration is constant a = −32 ft/. Use the general formulas for
motion with constant acceleration:
v(t) = at + v0 = −32t + 0 = −32t .
Now we use the velocity function and the one velocity value we know: v = −120
when it hits the ground. So the time when it hits the ground is given by
v(t) = −32t = −120 ⇒ t = 120/32 = 15/4
when it hits the ground. Now remember when it hits the ground the height is 0. So
s(15/4) = 0. But we know
s(t) = 12 at2 + v0 t + s0 = −16t2 + 0t + s0 = −16t2 + s0 .
Now substitute in t = 15/4 and solve for s0 .
s(15/4) = 0 ⇒ −16(15/4)2 + s0 = 0 ⇒ s0 = 152 = 225.
The cliff height is 225 feet.
EXAMPLE 6.11. A car is traveling at 90 km/h when the driver sees a deer 75 m ahead
and slams on the brakes. What constant deceleration is required to avoid hitting
Bambi? [Note: First convert 90 km/h to m/s.]
SOLUTION. Let’s list all that we know. v0 = 90 km/h or 90000
60·60 = 25 m/s and s0 = 0.
Let time t∗ represent the time it takes to stop. Then s(t∗ ) = 75 m. Now the car is
stopped at time t∗ , so we know v(t∗ ) = 0. Finally we know that acceleration is an
unknown constant, a, which is what we want to find.
Now we use our constant acceleration motion formulas to solve for a.
v(t) = at + v0 = at + 25
and
s(t) = 21 at2 + v0 t + s0 = 12 at2 + 25t.
Now use the other velocity and position we know: v(t∗ ) = 0 and s(t∗ ) = 75 when the
car stops. So
v(t∗ ) = at∗ + 25 = 0 ⇒ t∗ = −25/a
and
s(t∗ ) = 21 a(t∗ )2 + 25t∗ = 21 a(−25/a)2 + 25(−25/a) = 75.
Simplify to get
625a
625
625 1350
625
−
=
−
=−
= 75 ⇒ 150a = −625
a
2a
2a
2a
2a2
so
a=−
(Why is acceleration negative?)
625
25
=−
m/s.
150
6
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math 131, applications
motion: velocity and net change
6
EXAMPLE 6.12. One car intends to pass another on a back road. What constant acceleration is required to increase the speed of a car from 30 mph (44 ft/s) to 50 mph ( 220
3
ft/s) in 5 seconds?
SOLUTION. Given: a(t) = a constant. v0 = 44 ft/s. s0 = 0. And v(5) = 220
3 ft/s. Find
a. But
v(t) = at + v0 = at + 44.
So
v(5) = 5a + 44 =
Thus a =
220
220
88
⇒ 5a =
− 44 = .
3
3
3
88
15 .
YOU TRY IT 6.2. A toy bumper car is moving back and forth along a straight track. Its accel-
eration is a(t) = cos t + sin t. Find the particular velocity and position functions given that
v(π/4) = 0 and s(π ) = 1.
to you
R
R try it 6.2 : v(t) =
a(t) dt = cos t + sin t dt = sin t −
√
√
cos t + c. So v(π/4) = 22 − 22 + c =
0 ⇒ c = 0. Thus,
v(t) = Rsin t − cos t.
R
Now s(t) = v(t) dt = sin t −
cos t dt = − cos t − sin t + c. Since
s(π ) = −(−1) − 0 + c = 1 ⇒ c = 0. So
s(t) = − cos t − sin t.
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