12-3 Tangent Lines and Velocity
The position of an object in miles after t
minutes is given by s(t). Find the average
velocity of the object in miles per hour for the
given interval of time. Remember to convert
from minutes to hours.
18. for 3 ≤ t ≤ 5
The average velocity of the object is 0.75 miles per
minute. Thus, the average velocity of the object is 45
miles per hour.
21. s(t) = 0.01t3 − 0.01t2 for 4 ≤ t ≤ 7
SOLUTION: First, find the position of the object for a =4 and b =
7.
SOLUTION: First, find the position of the object for a =3 and b =
5.
Now use the formula for average velocity.
Now use the formula for average velocity.
The average velocity of the object is 0.75 miles per
minute. Thus, the average velocity of the object is 45
miles per hour.
21. s(t) = 0.01t3 − 0.01t2 for 4 ≤ t ≤ 7
0.82 · 60 = 49.2
The average velocity of the object is 0.82 miles per
minute. Thus, the average velocity of the object is
49.2 miles per hour.
24. TYPING The number of words w a person has
typed after t minutes is given by
SOLUTION: First, find the position of the object for a =4 and b =
7.
a. What was the average number of words per
minute the person typed between the 2nd and 4th
minutes?
b. What was the average number of words per
minute the person typed between the 3rd and 7th
minutes?
SOLUTION: a. First, find the number of words typed for a =2 and
b = 4.
Now use the formula for average velocity.
Now use the formula for average number of words.
0.82 · 60 = 49.2
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The average velocity of the object is 0.82 miles per
minute. Thus, the average velocity of the object is
49.2 miles per hour.
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The average number of words typed between the
2nd and 4th minutes is 46 words/min.
0.82 · 60 = 49.2
The average velocity of the object is 0.82 miles per
Thus, Lines
the average
of the object is
12-3minute.
Tangent
andvelocity
Velocity
49.2 miles per hour.
24. TYPING The number of words w a person has
typed after t minutes is given by
a. What was the average number of words per
minute the person typed between the 2nd and 4th
minutes?
b. What was the average number of words per
minute the person typed between the 3rd and 7th
minutes?
The average number of words typed between the
3rd and 7th minutes is 60.5 words/min.
The distance d an object is above the ground t
seconds after it is dropped is given by d(t). Find
the instantaneous velocity of the object at the
given value for t.
27. d(t) = −16t2 − 47t + 300; t = 1.5
SOLUTION: To find the instantaneous velocity, let t = 1.5 and
apply the formula for instantaneous velocity.
SOLUTION: a. First, find the number of words typed for a =2 and
b = 4.
Now use the formula for average number of words.
The instantaneous velocity of the object is –95 feet
per second.
30. d(t) = 150t − 16t2; t = 2.7
SOLUTION: The average number of words typed between the
2nd and 4th minutes is 46 words/min.
To find the instantaneous velocity, let t = 2.7 and
apply the formula for instantaneous velocity.
b. Find the number of words typed for a =3 and b =
7.
Now use the formula for average number of words.
The instantaneous velocity of the object is 63.6 feet
per second.
The average number of words typed between the
3rd and 7th minutes is 60.5 words/min.
The distance d an object is above the ground t
seconds after it is dropped is given by d(t). Find
the instantaneous velocity of the object at the
given value for t.
Find an equation for the instantaneous velocity
v(t) if the path of an object is defined as s(t) for
any point in time t.
33. s(t) = 14t2 − 7
SOLUTION: Apply the formula for instantaneous velocity.
27. d(t) = −16t2 − 47t + 300; t = 1.5
SOLUTION: eSolutions Manual - Powered by Cognero
To find the instantaneous velocity, let t = 1.5 and
apply the formula for instantaneous velocity.
Page 2
41. SKYDIVER Refer to the beginning of the lesson.
instantaneous
velocity
the object is 63.6 feet
12-3The
Tangent
Lines
andofVelocity
per second.
The position d of the skydiver in feet relative to the
2
ground can be defined by d(t) = 15,000 − 16t ,
where t is seconds passed after the skydiver exited
the plane.
Find an equation for the instantaneous velocity
v(t) if the path of an object is defined as s(t) for
any point in time t.
33. s(t) = 14t2 − 7
SOLUTION: Apply the formula for instantaneous velocity.
a. What is the average velocity of the skydiver
between the 2nd and 5th seconds of the jump?
b. What was the instantaneous velocity v0 of the
sky diver at 2 and 5 seconds?
c. Find an equation for the instantaneous velocity v
(t) of the skydiver.
SOLUTION: a. First, find the position of the skydiver for a =2 and
b = 5.
Now use the formula for average velocity.
The instantaneous velocity of the object at time t is v
(t) = 28t.
36. s(t) = 18 − t2 + 4t
SOLUTION: Apply the formula for instantaneous velocity.
The average velocity of the skydiver is −112 feet per
second.
b. Let t = 2 and apply the formula for instantaneous
velocity.
The instantaneous velocity of the object at time t is v
(t) = −2t + 4.
41. SKYDIVER Refer to the beginning of the lesson.
The position d of the skydiver in feet relative to the
2
ground can be defined by d(t) = 15,000 − 16t ,
where t is seconds passed after the skydiver exited
the plane.
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a. What is the average velocity of the skydiver
The instantaneous velocity of the object at t = 2 is
−64 feet per second.
Let t = 5 and apply the formula for instantaneous
velocity.
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The instantaneous velocity of the object at t = 2 is
−64 feet per second.
12-3 Tangent Lines and Velocity
Let t = 5 and apply the formula for instantaneous
velocity.
The instantaneous velocity of the object at t = 2 is
−160 feet per second.
c. Apply the formula for instantaneous velocity.
The instantaneous velocity of the object at time t is v
(t) = –32t.
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