An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Find the velocity function by differentiating the position function: v ( t ) = 104 − 32 t .
Determine when the object is at rest by setting v ( t ) = 0 , which gives t = 3.25 .
Find the intervals where the object moves up ( 0"> v ( t ) > 0 ) and down ( v ( t ) < 0 ), resulting in ( 0 , 3.25 ) and ( 3.25 , 6.5 ) , respectively.
Conclude that the object changes direction at t = 3.25 .
t = 3.25
Explanation
Problem Analysis We are given the position function s ( t ) = 104 t − 16 t 2 for an object moving along the s-axis, where 0 ≤ t ≤ 6.5 . We need to find when the object is at rest, when it moves up or down, and when it changes direction. To do this, we will analyze the velocity and acceleration functions.
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 The acceleration is constant and negative, which means the object is always decelerating.
Finding When the Object is at Rest To find when the object is momentarily at rest, we set the velocity function equal to zero and solve for t :
v ( t ) = 104 − 32 t = 0 32 t = 104 t = 32 104 = 3.25 So, the object is at rest when t = 3.25 seconds.
Determining When the Object Moves Up or Down Now, we determine when the object is moving up or down. The object moves upward when 0"> v ( t ) > 0 and downward when v ( t ) < 0 .
Since v ( t ) = 104 − 32 t , we have: 0"> v ( t ) > 0 when 0"> 104 − 32 t > 0 , which means t < 3.25 .
v ( t ) < 0 when 104 − 32 t < 0 , which means 3.25"> t > 3.25 .
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 the velocity changes sign. This occurs when v ( t ) = 0 , which we found to be at t = 3.25 seconds. Before this time, the object is moving up, and after this time, the object is moving down.
Final Answers a. The object is at rest when v ( t ) = 0 , which occurs at t = 3.25 seconds. b. 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 seconds.
Conclusion The object is at rest at t = 3.25 seconds. It moves upward from t = 0 to t = 3.25 seconds and downward from t = 3.25 to t = 6.5 seconds. The object changes direction at t = 3.25 seconds.
Examples
Understanding the motion of objects under constant acceleration is crucial in many real-world scenarios. For example, when designing amusement park rides, engineers need to calculate the velocity and acceleration of the carts to ensure the safety and enjoyment of the riders. Similarly, in sports, analyzing the trajectory of a ball or an athlete's movement involves understanding these concepts. By applying calculus to these situations, we can predict and control the motion of objects with precision.
