A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=-t^(3)+9t^(2)-24 t. Determine the speed of the particle at t=1. Answer:

Calculate Derivative of x(t): To find the speed of the particle at a specific time, we need to calculate the derivative of the position function x(t) with respect to time t, which gives us the velocity function v(t). The speed is the absolute value of velocity. Let's find the derivative of x(t) = -t^3 + 9t^2 - 24t. Using the power rule for derivatives, the derivative of t^n is n*t^(n-1).Find Velocity Function: Differentiate each term of x(t) separately: The derivative of -t^3 is -3t^2. The derivative of 9t^2 is 18t. The derivative of -24t is -24. So, the velocity function v(t) is v(t) = -3t^2 + 18t - 24.Evaluate Velocity at t=1: Now we need to evaluate the velocity function at t=1 to find the speed at that time. Substitute t=1 into v(t) = -3t^2 + 18t - 24. v(1) = -3(1)^2 + 18(1) - 24.Calculate Speed at t=1: Calculate the value of v(1): v(1) = -3(1) + 18(1) - 24. v(1) = -3 + 18 - 24. v(1) = 15 - 24. v(1) = -9.Final Result: The speed of the particle is the absolute value of the velocity. Speed at t=1 is |v(1)| = |-9|. Speed at t=1 is 9.

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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)+9t^(2)-24 t. Determine the speed of the particle at
t=1.
Answer:

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=-t^{3}+9 t^{2}-24 t$ . Determine the speed of the particle at $t=1$ . $\newline$ Answer:

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Math Problems

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}+9 t^{2}-24 t

. Determine the speed of the particle at

t=1

\newline

Answer:

Calculate Derivative of $x(t)$ : To find the speed of the particle at a specific time, we need to calculate the derivative of the position function $x(t)$ with respect to time $t$ , which gives us the velocity function $v(t)$ . The speed is the absolute value of velocity. $\newline$ Let's find the derivative of $x(t) = -t^3 + 9t^2 - 24t$ . $\newline$ Using the power rule for derivatives, the derivative of $t^n$ is $n\cdot t^{(n-1)}$ .
Find Velocity Function: Differentiate each term of $x(t)$ separately: $\newline$ The derivative of $-t^3$ is $-3t^2$ . $\newline$ The derivative of $9t^2$ is $18t$ . $\newline$ The derivative of $-24t$ is $-24$ . $\newline$ So, the velocity function $v(t)$ is $v(t) = -3t^2 + 18t - 24$ .
Evaluate Velocity at $t=1$ : Now we need to evaluate the velocity function at $t=1$ to find the speed at that time. $\newline$ Substitute $t=1$ into $v(t) = -3t^2 + 18t - 24$ . $\newline$ $v(1) = -3(1)^2 + 18(1) - 24$ .
Calculate Speed at $t=1$ : Calculate the value of $v(1)$ :
$v(1) = -3(1) + 18(1) - 24.$
$v(1) = -3 + 18 - 24.$
$v(1) = 15 - 24.$
$v(1) = -9.$
Final Result: The speed of the particle is the absolute value of the velocity. $\newline$ Speed at $t=1$ is $|v(1)| = |-9|$ . $\newline$ Speed at $t=1$ is $9$ .