A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=t^(3)-2t^(2)-39 t. Determine the velocity of the particle at t=4. 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 - 2t^2 - 39t. Let's find the derivative of this function, which will give us the velocity function v(t).Evaluate Velocity at t=4: Differentiate each term of x(t) with respect to t: The derivative of t^3 is 3t^2. The derivative of -2t^2 is -4t. The derivative of -39t is -39. So, the velocity function v(t) is v(t) = 3t^2 - 4t - 39.Calculate Velocity at t=4: Now we need to evaluate the velocity function at t=4 to find the velocity of the particle at that time. Substitute t with 4 in the velocity function v(t) = 3t^2 - 4t - 39. v(4) = 3(4)^2 - 4(4) - 39.Calculate the value of v(4): v(4) = 3(16) - 4(4) - 39, v(4) = 48 - 16 - 39, v(4) = 32 - 39, v(4) = -7.

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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)-2t^(2)-39 t. Determine the velocity of the particle at
t=4.
Answer:

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=t^{3}-2 t^{2}-39 t$ . Determine the velocity of the particle at $t=4$ . $\newline$ Answer:

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Calculus

Relate position, velocity, speed, and acceleration using derivatives

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Q. A particle moves along the

x

-axis so that at time

t \geq 0

its position is given by

x(t)=t^{3}-2 t^{2}-39 t

. Determine the velocity of the particle at

t=4

\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 $x(t) = t^3 - 2t^2 - 39t$ . Let's find the derivative of this function, which will give us the velocity function $v(t)$ .
Evaluate Velocity at $t=4$ : Differentiate each term of $x(t)$ with respect to $t$ :
The derivative of $t^3$ is $3t^2$ .
The derivative of $-2t^2$ is $-4t$ .
The derivative of $-39t$ is $-39$ .
So, the velocity function $v(t)$ is $x(t)$ $0$ .
Calculate Velocity at $t=4$ : Now we need to evaluate the velocity function at $t=4$ to find the velocity of the particle at that time. $\newline$ Substitute $t$ with $4$ in the velocity function $v(t) = 3t^2 - 4t - 39$ . $\newline$ $v(4) = 3(4)^2 - 4(4) - 39$ .
Calculate Velocity at $t=4$ : Now we need to evaluate the velocity function at $t=4$ to find the velocity of the particle at that time. $\newline$ Substitute $t$ with $4$ in the velocity function $v(t) = 3t^2 - 4t - 39$ . $\newline$ $v(4) = 3(4)^2 - 4(4) - 39$ .Calculate the value of $v(4)$ : $\newline$ $v(4) = 3(16) - 4(4) - 39$ , $\newline$ $v(4) = 48 - 16 - 39$ , $\newline$ $v(4) = 32 - 39$ , $\newline$ $t=4$ $0$ .