A particle is moving with acceleration [tex]$a(t)=24 t+2$[/tex]. Its position at time [tex]$t=0$[/tex] is [tex]$s(0)=5$[/tex] and its velocity at time [tex]$t=0$[/tex] is [tex]$v(0)=16$[/tex]. What is its position at time [tex]$t=6$[/tex]?
Answer (2)
By integrating the acceleration function twice and applying initial conditions, we find the position function of the particle. Evaluating this function at time t = 6 gives us the position as 1001 . Thus, the particle's position at t = 6 is 1001 .
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Integrate the acceleration function a ( t ) = 24 t + 2 to find the velocity function: v ( t ) = 12 t 2 + 2 t + C 1 , then use the initial condition v ( 0 ) = 16 to find C 1 = 16 , so v ( t ) = 12 t 2 + 2 t + 16 .
Integrate the velocity function v ( t ) = 12 t 2 + 2 t + 16 to find the position function: s ( t ) = 4 t 3 + t 2 + 16 t + C 2 , then use the initial condition s ( 0 ) = 5 to find C 2 = 5 , so s ( t ) = 4 t 3 + t 2 + 16 t + 5 .
Evaluate the position function at t = 6 : s ( 6 ) = 4 ( 6 ) 3 + ( 6 ) 2 + 16 ( 6 ) + 5 = 864 + 36 + 96 + 5 = 1001 .
The position of the particle at time t = 6 is 1001 .
Explanation
Problem Setup We are given the acceleration function a ( t ) = 24 t + 2 , the initial position s ( 0 ) = 5 , and the initial velocity v ( 0 ) = 16 . Our goal is to find the position of the particle at time t = 6 , which means we need to find s ( 6 ) .
Finding the Velocity Function First, we need to find the velocity function v ( t ) by integrating the acceleration function a ( t ) with respect to time t :
v(t) = \[ \int a(t) dt = \int (24t + 2) dt \]
Integrating Acceleration Integrating a ( t ) , we get: v ( t ) = 12 t 2 + 2 t + C 1 where C 1 is the constant of integration.
Using Initial Velocity We use the initial condition v ( 0 ) = 16 to find C 1 :
16 = 12 ( 0 ) 2 + 2 ( 0 ) + C 1 C 1 = 16 So, the velocity function is: v ( t ) = 12 t 2 + 2 t + 16
Finding the Position Function Next, we need to find the position function s ( t ) by integrating the velocity function v ( t ) with respect to time t :
s ( t ) = ∫ v ( t ) d t = ∫ ( 12 t 2 + 2 t + 16 ) d t
Integrating Velocity Integrating v ( t ) , we get: s ( t ) = 4 t 3 + t 2 + 16 t + C 2 where C 2 is the constant of integration.
Using Initial Position We use the initial condition s ( 0 ) = 5 to find C 2 :
5 = 4 ( 0 ) 3 + ( 0 ) 2 + 16 ( 0 ) + C 2 C 2 = 5 So, the position function is: s ( t ) = 4 t 3 + t 2 + 16 t + 5
Evaluating Position at t=6 Finally, we evaluate the position function at t = 6 to find the position of the particle at t = 6 :
s ( 6 ) = 4 ( 6 ) 3 + ( 6 ) 2 + 16 ( 6 ) + 5 s ( 6 ) = 4 ( 216 ) + 36 + 96 + 5 s ( 6 ) = 864 + 36 + 96 + 5 s ( 6 ) = 1001
Final Answer The position of the particle at time t = 6 is 1001.
Examples
Understanding motion is crucial in many fields. For example, in physics, knowing the acceleration, initial velocity, and initial position of an object allows us to predict its future location. This is used in projectile motion calculations, satellite trajectory predictions, and even in designing safer cars by analyzing collision dynamics. By using calculus to relate acceleration, velocity, and position, we can model and understand the world around us more accurately.
