The function [tex]$s=f(t)$[/tex] gives the position of an object moving along the [tex]$s$[/tex]-axis as a function of time [tex]$t$[/tex]. Graph [tex]$f$[/tex] together with the velocity function [tex]$v(t)=\frac{d s}{d t}=f(t)$[/tex] and the acceleration function [tex]$a(t)=\frac{d^2 s}{d^2}=t^{\prime \prime}(t)$[/tex], then complete parts (a) through (f).
[tex]$s=104 t-16 t^2, 0 \leq t \leq 6.5$[/tex] (a heavy object fired straight up from Earth's surface at [tex]$104 f / sec$[/tex])
A. The object is moving down for [tex]$t$[/tex] in the interval ______ but is never moving up.
B. The object is moving down for [tex]$t$[/tex] in the interval ([tex]$3.25,6.5$[/tex]] and the object is moving up for [tex]$t$[/tex] in the interval [[tex]$0,3.25$[/tex])
C. The object is never moving down, but is moving up for [tex]$t$[/tex] in the interval ______.
D. The object is never moving down or up.
c. When does the object change direction? Select the correct answer below, and if necessary, fill in the answer box to complete your choice.
A. The object changes direction at [tex]$t =3.25 sec$[/tex].
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B. The object never changes direction.
d. When does the object speed up and slow down? Select the correct answer below, and if necessary, till in the answer box(es) to complete your choice. (Simplify your answer. Type your answer in interval notation. Use integers or decimals for any numbers in the expression)
A. The object never speeds up, but slows down for [tex]$t$[/tex] in the interval ______.
B. The object speeds up for [tex]$t$[/tex] in the interval ([tex]$3.25,6.5$[/tex]) and the object slows down for [tex]$t$[/tex] in the intervat ([tex]$0,3.25$[/tex])
C. The object speeds up for [tex]$t$[/tex] in the interval ______ but never slows down.
D. The object never speeds up or slows down.
e. When is the object moving fastest (highest speed)? Stowest? Select the correct answer below and till in any answer boxes to complete your choice.
A. The object is moving fastest at [tex]$t =$[/tex] ______ when the speed is ______ fi/sec. The object is moving slowest at [tex]$t =$[/tex] ______ when the speed is ______ [tex]$n / sec$[/tex]
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B. The object is moving at a constant speed of ______ [tex]$fl / sec$[/tex]
Answer (1)
t = 3.25
Explanation
Problem Setup We are given the position function s ( t ) = 104 t − 16 t 2 for an object moving along the s-axis, with 0 l e t l e 6.5 . We need to analyze the motion of the object, including when it moves up/down, changes direction, speeds up/slows down, and when it is moving fastest/slowest. To do this, we will find the velocity and acceleration functions.
Finding Velocity Function First, we find the velocity function v ( t ) by taking the derivative of s ( t ) with respect to t :
v ( t ) = d t d ( 104 t − 16 t 2 ) = 104 − 32 t
Finding Acceleration Function Next, we find the acceleration function a ( t ) by taking the derivative of v ( t ) with respect to t :
a ( t ) = d t d ( 104 − 32 t ) = − 32
Determining Upward and Downward Motion Now, we determine when the object is moving up and down. The object is moving up when 0"> v ( t ) > 0 and moving down when v ( t ) < 0 . We find when v ( t ) = 0 :
104 − 32 t = 0 ⇒ t = 32 104 = 3.25 So, 0"> v ( t ) > 0 when 0 l e t < 3.25 and v ( t ) < 0 when 3.25 < t l e 6.5 . Therefore, the object is moving up for t in the interval [ 0 , 3.25 ) and moving down for t in the interval ( 3.25 , 6.5 ] .
Finding When the Object Changes Direction The object changes direction when v ( t ) = 0 and the sign of v ( t ) changes. This occurs at t = 3.25 .
Determining When the Object Speeds Up and Slows Down To determine when the object is speeding up and slowing down, we look at the signs of v ( t ) and a ( t ) .
Speeding up: v ( t ) and a ( t ) have the same sign.
Slowing down: v ( t ) and a ( t ) have opposite signs. Since a ( t ) = − 32 is always negative, the object is speeding up when v ( t ) < 0 and slowing down when 0"> v ( t ) > 0 .
Speeding up: 3.25 < t l e 6.5
Slowing down: 0 l e t < 3.25 Thus, the object speeds up for t in the interval ( 3.25 , 6.5 ] and slows down for t in the interval [ 0 , 3.25 ) .
Finding Fastest and Slowest Speeds To find when the object is moving fastest and slowest, we need to find the maximum and minimum values of ∣ v ( t ) ∣ on the interval [ 0 , 6.5 ] .
At t = 0 , ∣ v ( 0 ) ∣ = ∣104 − 32 ( 0 ) ∣ = 104 ft/sec.
At t = 3.25 , ∣ v ( 3.25 ) ∣ = ∣104 − 32 ( 3.25 ) ∣ = 0 ft/sec.
At t = 6.5 , ∣ v ( 6.5 ) ∣ = ∣104 − 32 ( 6.5 ) ∣ = ∣ − 104∣ = 104 ft/sec. The object is moving fastest at t = 0 and t = 6.5 , when the speed is 104 ft/sec. The object is moving slowest at t = 3.25 , when the speed is 0 ft/sec.
Final Answers Based on our analysis: A. The object is moving down for t in the interval ( 3.25 , 6.5 ] and the object is moving up for t in the interval [ 0 , 3.25 ) .
c. The object changes direction at t = 3.25 sec. d. The object speeds up for t in the interval ( 3.25 , 6.5 ] and the object slows down for t in the interval [ 0 , 3.25 ) .
e. The object is moving fastest at t = 0 and t = 6.5 when the speed is 104 ft/sec. The object is moving slowest at t = 3.25 when the speed is 0 ft/sec.
Summary of Motion The object's motion is described by its position, velocity, and acceleration functions. Understanding these concepts allows us to analyze how the object moves, changes direction, and varies its speed over time.
Examples
Imagine you're designing a roller coaster. The height of the coaster at any point can be modeled by a position function, similar to the one in this problem. By analyzing the velocity and acceleration, you can determine where the coaster will reach its maximum speed, where it will slow down, and where it will change direction, ensuring a thrilling yet safe ride. Understanding these calculus concepts is crucial for designing a safe and exciting roller coaster experience.
