The position function of a particle is given by [tex]$s=t^3-10.5 t^2-2 t, t \geq 0$[/tex]. When does the particle reach a velocity of [tex]$52 M / s$[/tex]?
The particle reaches a velocity of [tex]$52 M / s$[/tex] when [tex]$t=$[/tex] _____ seconds.
Answer (1)
Find the velocity function by taking the derivative of the position function: v ( t ) = 3 t 2 − 21 t − 2 .
Set the velocity function equal to 52: 3 t 2 − 21 t − 2 = 52 , which simplifies to 3 t 2 − 21 t − 54 = 0 .
Solve the quadratic equation for t , obtaining t = − 2 and t = 9 .
Since t ≥ 0 , the particle reaches a velocity of 52 m / s when t = 9 seconds.
Explanation
Finding the Velocity Function We are given the position function s ( t ) = t 3 − 10.5 t 2 − 2 t and we want to find the time t when the velocity is 52 m / s . To do this, we first need to find the velocity function v ( t ) , which is the derivative of the position function with respect to time.
Calculating the Derivative The velocity function v ( t ) is the derivative of the position function s ( t ) :
v ( t ) = d t d s = 3 t 2 − 21 t − 2
Setting up the Equation Now, we set the velocity function equal to 52 and solve for t :
3 t 2 − 21 t − 2 = 52
3 t 2 − 21 t − 54 = 0
Solving the Quadratic Equation We can solve this quadratic equation for t . The quadratic equation is in the form a t 2 + b t + c = 0 , where a = 3 , b = − 21 , and c = − 54 . We can use the quadratic formula to find the roots, but in this case, we will use python tool to find the roots:
The roots are t = − 2 and t = 9 .
Determining the Valid Solution Since time t must be greater than or equal to 0 , we discard the negative solution t = − 2 . Therefore, the particle reaches a velocity of 52 m / s when t = 9 seconds.
Final Answer Thus, the particle reaches a velocity of 52 m / s at t = 9 seconds.
Examples
Understanding particle motion is crucial in physics and engineering. For example, when designing a roller coaster, engineers use position, velocity, and acceleration functions to ensure the ride is thrilling but safe. By calculating the velocity at different points, they can optimize the track design to provide the desired experience while adhering to safety regulations. This blend of mathematical modeling and practical application ensures an exciting and secure ride.
