The function [tex]s = 104 - te^2, 0 \leq I \leq 6.5[/tex] gives the position of an object moving along the s-axis as a function of time [tex]t[/tex]. Graph this together with the velocity function [tex]v(t) = \frac{ds}{dt} = f'(t)[/tex] and the acceleration function [tex]a(t) = \frac{d^2 s}{dt^2} = v'(t)[/tex], then complete parts (a) through (f).
(a heavy object fired straight up from Earth's surface at 104 ft/sec)
[tex]v(t) = 104 - 32t[/tex]
[tex]a(t) = -32[/tex]
Choose the correct graph of [tex]s = f(t)[/tex], [tex]v(t)[/tex], and [tex]a(t)[/tex] between. The window for each graph is [0, 6.5] by [-110, 250].
A.
B.
C.
D.
a. When is the object momentarily at rest? Select the correct answer below, and if necessary, fill in the answer box(es) to complete your choice.
A. The object is at rest when velocity = 0 ft/sec, and this occurs if [tex]t =[/tex] (Type integers or decimals. Use a comma to separate answers as needed.)
B. The object is never at rest.
b. When does it move down or up? 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 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 and the object is moving up for [tex]t[/tex] in the interval
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.
Answer (1)
Correct the position, velocity, and acceleration functions based on the problem description.
Find the time when the object is at rest by setting the velocity function to zero: v ( t ) = 104 − 32 t = 0 , which gives t = 3.25 .
Determine the intervals for upward and downward motion by analyzing the sign of the velocity function: upward when 0"> v ( t ) > 0 and downward when v ( t ) < 0 .
The object is moving up for t in the interval [ 0 , 3.25 ) and down for t in the interval ( 3.25 , 6.5 ] .
Explanation
Problem Analysis The problem describes the motion of an object along the s-axis and asks us to analyze its position, velocity, and acceleration. We are given the functions for position s ( t ) , velocity v ( t ) , and acceleration a ( t ) , although the initial expressions for s ( t ) and v ( t ) appear to be incorrect. We need to correct these functions, find when the object is at rest, and determine the intervals when the object is moving up or down.
Correcting the Functions First, let's correct the given functions. Assuming the object is fired straight up from Earth's surface with an initial velocity, the position function should be of the form s ( t ) = v 0 t + 2 1 a t 2 , where v 0 is the initial velocity and a is the acceleration due to gravity. Given that a ( t ) = − 32 (which is correct), the velocity function should be the integral of the acceleration, v ( t ) = ∫ a ( t ) d t = ∫ − 32 d t = − 32 t + C . Since the initial velocity is given as 104, we have v ( 0 ) = 104 , so C = 104 . Thus, v ( t ) = 104 − 32 t . The position function is the integral of the velocity function, s ( t ) = ∫ v ( t ) d t = ∫ ( 104 − 32 t ) d t = 104 t − 16 t 2 + D . Assuming the initial position is 0, we have s ( 0 ) = 0 , so D = 0 . Thus, s ( t ) = 104 t − 16 t 2 .
Finding When the Object is at Rest Now, let's find when the object is momentarily at rest. This occurs when the velocity is zero, i.e., v ( t ) = 0 . So, we solve 104 − 32 t = 0 for t : 104 − 32 t = 0 32 t = 104 t = 32 104 = 4 13 = 3.25 So, the object is at rest when t = 3.25 .
Determining When the Object is Moving Up or Down Next, let's determine when the object is moving up or down. The object is moving up when 0"> v ( t ) > 0 and moving down when v ( t ) < 0 . For 0"> v ( t ) > 0 :
0"> 104 − 32 t > 0 32 t < 104 t < 32 104 = 3.25 So, the object is moving up when 0 ≤ t < 3.25 .
For v ( t ) < 0 :
104 − 32 t < 0 104"> 32 t > 104 \frac{104}{32} = 3.25"> t > 32 104 = 3.25 So, the object is moving down when 3.25 < t ≤ 6.5 .
Conclusion Based on the analysis:
The object is at rest when t = 3.25 .
The object is moving up for t in the interval [ 0 , 3.25 ) .
The object is moving down for t in the interval ( 3.25 , 6.5 ] .
Examples
Understanding projectile motion is crucial in various fields, such as sports and engineering. For example, when designing a catapult or launching a rocket, engineers need to calculate the trajectory, velocity, and acceleration of the projectile to ensure it reaches its target accurately. By applying the principles of calculus and physics, they can optimize the launch angle and initial velocity to achieve the desired range and height. This problem demonstrates the fundamental concepts used in these real-world applications.
