A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=t^(3)-12t^(2)+21 t. Determine the velocity of the particle at t=6. Answer:

Derivative of Position Function: To find the velocity of the particle at a specific time, we need to take the derivative of the position function with respect to time, because velocity is the rate of change of position.Velocity Function Derivation: The position function is x(t) = t^3 - 12t^2 + 21t. Let's find the derivative of this function, which will give us the velocity function v(t).Evaluate Velocity at t=6: Differentiate each term of x(t) with respect to t: The derivative of t^3 is 3t^2. The derivative of -12t^2 is -24t. The derivative of 21t is 21. So, the velocity function v(t) is v(t) = 3t^2 - 24t + 21.Calculate Velocity at t=6: Now we need to evaluate the velocity function at t=6 to find the velocity of the particle at that time. Substitute t with 6 in the velocity function v(t) = 3t^2 - 24t + 21. v(6) = 3(6)^2 - 24(6) + 21.Calculate the value of v(6): v(6) = 3(36) - 24(6) + 21 v(6) = 108 - 144 + 21 v(6) = -36 + 21 v(6) = -15

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A particle moves along the
x-axis so that at time
t >= 0 its position is given by
x(t)=t^(3)-12t^(2)+21 t. Determine the velocity of the particle at
t=6.
Answer:

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=t^{3}-12 t^{2}+21 t$ . Determine the velocity of the particle at $t=6$ . $\newline$ Answer:

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Calculus

Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle moves along the

x

-axis so that at time

t \geq 0

its position is given by

x(t)=t^{3}-12 t^{2}+21 t

. Determine the velocity of the particle at

t=6

\newline

Answer:

Derivative of Position Function: To find the velocity of the particle at a specific time, we need to take the derivative of the position function with respect to time, because velocity is the rate of change of position.
Velocity Function Derivation: The position function is $x(t) = t^3 - 12t^2 + 21t$ . Let's find the derivative of this function, which will give us the velocity function $v(t)$ .
Evaluate Velocity at $t=6$ : Differentiate each term of $x(t)$ with respect to $t$ :
The derivative of $t^3$ is $3t^2$ .
The derivative of $-12t^2$ is $-24t$ .
The derivative of $21t$ is $21$ .
So, the velocity function $v(t)$ is $x(t)$ $0$ .
Calculate Velocity at $t=6$ : Now we need to evaluate the velocity function at $t=6$ to find the velocity of the particle at that time. $\newline$ Substitute $t$ with $6$ in the velocity function $v(t) = 3t^2 - 24t + 21$ . $\newline$ $v(6) = 3(6)^2 - 24(6) + 21$ .
Calculate Velocity at $t=6$ : Now we need to evaluate the velocity function at $t=6$ to find the velocity of the particle at that time. $\newline$ Substitute $t$ with $6$ in the velocity function $v(t) = 3t^2 - 24t + 21$ . $\newline$ $v(6) = 3(6)^2 - 24(6) + 21$ .Calculate the value of $v(6)$ : $\newline$ $v(6) = 3(36) - 24(6) + 21$ $\newline$ $v(6) = 108 - 144 + 21$ $\newline$ $v(6) = -36 + 21$ $\newline$ $t=6$ $0$