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])
a. The object is moving down for [tex]$t$[/tex] in the interval ______ but is never moving up.
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 (3.25, 6.5) and the object is moving up for [tex]$t$[/tex] in the interval [0, 3.25).
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]$t = 3.25$[/tex] sec.
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.
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 (3.25, 6.5] and the object slows down for [tex]$t$[/tex] in the interval [0, 3.25).
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]$t =$[/tex] ______ when the speed is ______ ft/sec. The object is moving slowest at [tex]$t =$[/tex] ______ when the speed is ______ ft/sec
B. The object is moving at a constant speed of ______ ft/sec.
Answer (2)
( 3.25 , 6.5 ] , 3.25 , ( 3.25 , 6.5 ] , [ 0 , 3.25 ) , 0 , 104 , 3.25 , 0
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, with 0 l e t g t r 6.5 . We need to analyze the motion of this object.
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 s = 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 v = − 32
Determining When Object Moves Up Now, we determine when the object is moving up. The object is moving up when 0"> v ( t ) > 0 . So, we solve the inequality 0"> 104 − 32 t > 0 :
32t"> 104 > 32 t t < 32 104 = 3.25 Thus, the object is moving up for 0 g t r t < 3.25 .
Determining When Object Moves Down Similarly, the object is moving down when v ( t ) < 0 . So, we solve the inequality 104 − 32 t < 0 :
104 < 32 t \frac{104}{32} = 3.25"> t > 32 104 = 3.25 Thus, the object is moving down for 3.25 < t g t r 6.5 .
Determining When Object Changes Direction The object changes direction when v ( t ) = 0 . We already found that v ( t ) = 0 when t = 3.25 .
Determining When Object Speeds Up The object speeds up when v ( t ) and a ( t ) have the same sign. Since a ( t ) = − 32 is always negative, the object speeds up when v ( t ) < 0 , which is when 3.25"> t > 3.25 . So, the object speeds up for 3.25 < t g t r 6.5 .
Determining When Object Slows Down The object slows down when v ( t ) and a ( t ) have opposite signs. Since a ( t ) = − 32 is always negative, the object slows down when 0"> v ( t ) > 0 , which is when t < 3.25 . So, the object slows down for 0 g t r t < 3.25 .
Determining When Object Moves Fastest and Slowest The object is moving fastest when the absolute value of its velocity is greatest. Since v ( t ) is a linear function, the maximum speed occurs at the endpoints of the interval [ 0 , 6.5 ] .
At t = 0 , v ( 0 ) = 104 − 32 ( 0 ) = 104 ft/sec. At t = 6.5 , v ( 6.5 ) = 104 − 32 ( 6.5 ) = 104 − 208 = − 104 ft/sec. So, the object is moving fastest at t = 0 and t = 6.5 , when the speed is 104 ft/sec. The object is moving slowest when its speed is zero, which occurs at t = 3.25 when the speed is 0 ft/sec.
Final Answers Based on the above analysis: A. The object is moving down for t in the interval ( 3.25 , 6.5 ) .
B. The object changes direction at t = 3.25 sec. D. The object speeds up for t in the interval ( 3.25 , 6.5 ] and slows down for t in the interval [ 0 , 3.25 ) .
E. The object is moving fastest at t = 0 when the speed is 104 ft/sec. The object is moving slowest at t = 3.25 when the speed is 0 ft/sec.
Examples
Understanding projectile motion is crucial in many fields. For example, in sports, knowing the launch angle and initial velocity of a ball helps athletes optimize their performance. Similarly, engineers use these principles to design safer and more efficient transportation systems, such as calculating the trajectory of a rocket or the braking distance of a car. By analyzing the position, velocity, and acceleration of objects, we can make accurate predictions and improve various aspects of our lives.
The object moves down in the interval (3.25, 6.5), changes direction at t = 3.25 sec, speeds up for t > 3.25, slows down for t < 3.25, moves fastest at t = 0 with 104 ft/sec, and slowest at t = 3.25 with 0 ft/sec.
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