A particle moves along a straight line relative to a fixed origin [tex]$O$[/tex] with acceleration, measured in [tex]$m / s ^2$[/tex], given by the function [tex]$a(t)=8 t^3-6 t$[/tex], where [tex]$t$[/tex] is the time in seconds. If the particle has velocity [tex]$v=-4 m / s$[/tex] when [tex]$t=2 s$[/tex], find the velocity (in [tex]$m / s$[/tex]) of the particle at time [tex]$t$[/tex].
a [tex]$v(t)=2 t^4-3 t^2-8$[/tex]
b [tex]$v(t)=2 t^4-3 t^2-24$[/tex]
c [tex]$v(t)=t^4-2 t^2-12$[/tex]
d [tex]$v(t)=4 t^3-3 t-30$[/tex]
e [tex]$v(t)=4 t^4-3 t^2-56$[/tex]
Answer (2)
Integrate the acceleration function a ( t ) = 8 t 3 − 6 t to find the velocity function: v ( t ) = 2 t 4 − 3 t 2 + C .
Use the initial condition v ( 2 ) = − 4 to solve for the constant of integration C .
Substitute t = 2 into the velocity function: − 4 = 2 ( 2 ) 4 − 3 ( 2 ) 2 + C , which simplifies to − 4 = 32 − 12 + C .
Solve for C to find C = − 24 , so the velocity function is v ( t ) = 2 t 4 − 3 t 2 − 24 .
Explanation
Problem Setup We are given the acceleration function a ( t ) = 8 t 3 − 6 t and the initial condition v ( 2 ) = − 4 m/s. Our goal is to find the velocity function v ( t ) .
Integrating Acceleration To find the velocity function, we need to integrate the acceleration function with respect to time t :
v(t) = \[ \int a(t) dt = \int (8t^3 - 6t) dt \]
Velocity Function with Constant Integrating the acceleration function, we get: v ( t ) = 2 t 4 − 3 t 2 + C where C is the constant of integration.
Applying Initial Condition Now, we use the initial condition v ( 2 ) = − 4 to find the value of C . Substituting t = 2 into the velocity function, we have: − 4 = 2 ( 2 ) 4 − 3 ( 2 ) 2 + C − 4 = 2 ( 16 ) − 3 ( 4 ) + C − 4 = 32 − 12 + C − 4 = 20 + C
Solving for Constant Solving for C , we get: C = − 4 − 20 = − 24
Final Velocity Function Therefore, the velocity function is: v ( t ) = 2 t 4 − 3 t 2 − 24
Examples
Understanding the motion of objects is crucial in many fields. For example, in physics, knowing the acceleration of a rocket allows us to determine its velocity and position at any time. This is essential for trajectory planning and control. Similarly, in engineering, analyzing the acceleration of a car helps in designing safer and more efficient braking systems. The relationship between acceleration, velocity, and position is fundamental in understanding and predicting the behavior of moving objects.
To find the particle's velocity, we integrated the acceleration function a ( t ) = 8 t 3 − 6 t , resulting in v ( t ) = 2 t 4 − 3 t 2 + C . Using the condition v ( 2 ) = − 4 , we found C = − 24 , giving us the velocity function v ( t ) = 2 t 4 − 3 t 2 − 24 . Thus, the correct answer is option b: v ( t ) = 2 t 4 − 3 t 2 − 24 .
;
