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]$v(t)=\frac{d s}{d t}=f^{\prime}(t)$[/tex] and the acceleration function [tex]$a(t)=\frac{d^2 s}{d t^2}=t^{\prime \prime}(t)$[/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])
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, fill 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 interval [[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)? Slowest? Select the correct answer below and fill in any answer boxes to complete your choice.
A. The object is moving fastest at [tex]$t=0,6.5$[/tex], when the speed is [tex]$104 f / sec$[/tex]. The object is moving slowest at [tex]$t=3.25$[/tex], when the speed is [tex]$0 ft / sec$[/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 ______ f I/sec.
f. When is the object farthest from the axis origin?
When [tex]$t =$[/tex] ______ sec, the object is ______ feet from the origin, which is the farthest away.
Answer (2)
Determine the velocity function by differentiating the position function: v ( t ) = 104 − 32 t .
Find when the object changes direction by setting v ( t ) = 0 , which gives t = 3.25 seconds.
Analyze the signs of v ( t ) and a ( t ) to determine when the object speeds up (when v ( t ) and a ( t ) have the same sign) and slows down (when v ( t ) and a ( t ) have opposite signs).
Find the maximum distance from the origin by evaluating s ( t ) at t = 3.25 , resulting in a distance of 169 feet.
Explanation
Problem Setup We are given the position function s ( t ) = 104 t − 16 t 2 for an object fired straight up from Earth's surface at 104 ft/sec, where $0 \le t le 6.5$. We need to analyze the motion of this object.
Finding Velocity and Acceleration First, we find the velocity function v ( t ) by taking the derivative of the position function s ( t ) with respect to time t : v ( t ) = d t d s = 104 − 32 t Next, we find the acceleration function a ( t ) by taking the derivative of the velocity function v ( t ) with respect to time t : a ( t ) = d t d v = − 32
Determining Direction of Motion To determine when the object is moving up or down, we analyze the sign of the velocity function. The object is moving up when 0"> v ( t ) > 0 and moving down when v ( t ) < 0 . 0 \Rightarrow t < \frac{104}{32} = 3.25"> 104 − 32 t > 0 ⇒ t < 32 104 = 3.25 \frac{104}{32} = 3.25"> 104 − 32 t < 0 ⇒ t > 32 104 = 3.25 Thus, 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 . 104 − 32 t = 0 ⇒ t = 32 104 = 3.25 So, the object changes direction at t = 3.25 sec.
Determining When the Object Speeds Up and Slows Down To determine when the object is speeding up or slowing down, we analyze the signs of the velocity and acceleration functions. Since a ( t ) = − 32 is always negative, the object speeds up when v ( t ) < 0 and slows down when 0"> v ( t ) > 0 .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 When the Object is Moving Fastest and Slowest To find when the object is moving fastest, we look for the maximum value of ∣ v ( t ) ∣ on the interval [ 0 , 6.5 ] . ∣ v ( 0 ) ∣ = ∣104 − 32 ( 0 ) ∣ = 104 ∣ v ( 6.5 ) ∣ = ∣104 − 32 ( 6.5 ) ∣ = ∣ − 104∣ = 104 The object is moving slowest when v ( t ) = 0 , which occurs at t = 3.25 . The speed at this time is 0 ft/sec.Thus, 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.
Finding When the Object is Farthest from the Origin To find when the object is farthest from the origin, we look for the maximum value of s ( t ) on the interval [ 0 , 6.5 ] . This occurs when v ( t ) = 0 , which is at t = 3.25 . s ( 3.25 ) = 104 ( 3.25 ) − 16 ( 3.25 ) 2 = 338 − 16 ( 10.5625 ) = 338 − 169 = 169 When t = 3.25 sec, the object is 169 feet from the origin, which is the farthest away.
Final Answers Based on our analysis:
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 ) .
The object changes direction at t = 3.25 sec.
The object speeds up for t in the interval ( 3.25 , 6.5 ] and slows down for t in the interval [ 0 , 3.25 ) .
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.
When t = 3.25 sec, the object is 169 feet from the origin, which is the farthest away.
Examples
Understanding projectile motion is crucial in fields like sports and engineering. For example, when designing a catapult or analyzing a baseball trajectory, knowing the initial velocity, launch angle, and the effect of gravity allows engineers and athletes to predict the range, maximum height, and flight time of the projectile. This problem demonstrates how calculus can be used to model and analyze such scenarios, optimizing performance and safety.
The object is moving up in the interval [0, 3.25) and down in the interval (3.25, 6.5]. It changes direction at t = 3.25 seconds and speeds up in the interval (3.25, 6.5] while slowing down in [0, 3.25). The object reaches its maximum height of 169 feet at t = 3.25 seconds.
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