The function $s=f(t)$ gives the position of an object moving along the $s$-axis as a function of time $t$. Graph $f$ together with the velocity function $v(t)=\frac{d s}{d t}=f^{\prime}(t)$ and the acceleration function $a(t)=\frac{d^2 s}{d t^2}=f^{\prime \prime}(t)$, then complete parts (a) through (f).
$s =104 t -16 t ^2, 0 \leq t \leq 6.5$ (a heavy object fired straight up from Earth's surface at $104 ft / sec$ )

$v(t)=104-32 t $
$a(t)=-32$

Choose the correct graph of $s=f(t), v(t)$, and a(t) below. The window for each graph is $[0,6.5]$ by $[-110,250]$.
A. B. c.
D.
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 $v(t)=\square f / sec$, and this occurs at $t=\square sec$.

(Type integers or decimals. Use a comma to separate answers as needed.)
B. The object is never at rest.

The object is at rest when its velocity is zero, so set v ( t ) = 104 − 32 t = 0 .
Solve for t : 32 t = 104 , which gives t = 32 104 ​ = 3.25 .
The object is at rest when v ( t ) = 0 ft/sec, and this occurs at t = 3.25 sec.
The correct answer is A, and the object is at rest at t = 3.25 ​ .

Explanation

Problem Analysis We are given the position function s ( t ) = 104 t − 16 t 2 , the velocity function v ( t ) = 104 − 32 t , and the acceleration function a ( t ) = − 32 . We need to find when the object is momentarily at rest and choose the correct graph.

Finding Time at Rest The object is at rest when its velocity is zero. So, we need to solve the equation v ( t ) = 0 for t .

Solving for t We have 104 − 32 t = 0 . Adding 32 t to both sides gives 104 = 32 t . Dividing both sides by 32 gives t = 32 104 ​ = 4 13 ​ = 3.25 $.

Velocity at t=3.25 So, the object is at rest at t = 3.25 seconds. The velocity at this time is v ( 3.25 ) = 104 − 32 ( 3.25 ) = 104 − 104 = 0 ft/sec.

Analyzing the Graphs Now, let's analyze the graphs. The position function s ( t ) = 104 t − 16 t 2 is a parabola opening downwards. The vertex of the parabola is at t = 3.25 , and the value of s ( 3.25 ) is 104 ( 3.25 ) − 16 ( 3.25 ) 2 = 338 − 16 ( 10.5625 ) = 338 − 169 = 169 . So the vertex is at ( 3.25 , 169 ) . The velocity function v ( t ) = 104 − 32 t is a straight line with a negative slope, and it crosses the t-axis at t = 3.25 . The acceleration function a ( t ) = − 32 is a horizontal line at y = − 32 . Based on these characteristics, graph A is the correct one.

Final Answer Therefore, the object is at rest when v ( t ) = 0 ft/sec, and this occurs at t = 3.25 sec.

Choosing the Correct Graph The correct graph is A.


Examples
Understanding when an object is at rest is crucial in physics and engineering. For example, when designing a roller coaster, engineers need to calculate the velocity and acceleration at different points to ensure the ride is safe and thrilling. Knowing when the coaster momentarily stops or changes direction helps in designing the track and ensuring passenger safety.

Answered by GinnyAnswer | 2025-07-07