A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=-t^(3)+10t^(2)-25 t. Determine the acceleration of the particle at t=1. Answer:

Find Acceleration Function: To find the acceleration of the particle at a specific time, we need to find the second derivative of the position function x(t) with respect to time t. The first derivative of x(t) with respect to t gives us the velocity v(t), and the second derivative gives us the acceleration a(t).Find Velocity Function: First, we find the first derivative of x(t) which is the velocity function v(t). The derivative of -t^(3) is -3t^(2), the derivative of 10t^(2) is 20t, and the derivative of -25t is -25. So, v(t) = -3t^(2) + 20t - 25.Evaluate Acceleration at t=1: Next, we find the second derivative of x(t) which is the acceleration function a(t). The derivative of -3t^(2) is -6t, the derivative of 20t is 20, and the derivative of a constant (-25) is 0. So, a(t) = -6t + 20.Calculate Acceleration at t=1: Now, we evaluate the acceleration function a(t) at t=1 to find the acceleration of the particle at that time. Substituting t=1 into a(t) gives us a(1) = -6(1) + 20.Calculating a(1) gives us -6 + 20, which equals 14. Therefore, the acceleration of the particle at t=1 is 14 meters per second squared.

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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)+10t^(2)-25 t. Determine the acceleration 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 $x(t)=-t^{3}+10 t^{2}-25 t$ . Determine the acceleration of the particle at $t=1$ . $\newline$ 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}+10 t^{2}-25 t

. Determine the acceleration of the particle at

t=1

\newline

Answer:

Find Acceleration Function: To find the acceleration of the particle at a specific time, we need to find the second derivative of the position function $x(t)$ with respect to time $t$ . The first derivative of $x(t)$ with respect to $t$ gives us the velocity $v(t)$ , and the second derivative gives us the acceleration $a(t)$ .
Find Velocity Function: First, we find the first derivative of $x(t)$ which is the velocity function $v(t)$ . The derivative of $-t^{3}$ is $-3t^{2}$ , the derivative of $10t^{2}$ is $20t$ , and the derivative of $-25t$ is $-25$ . So, $v(t) = -3t^{2} + 20t - 25$ .
Evaluate Acceleration at $t=1$ : Next, we find the second derivative of $x(t)$ which is the acceleration function $a(t)$ . The derivative of $-3t^{2}$ is $-6t$ , the derivative of $20t$ is $20$ , and the derivative of a constant $(-25)$ is $0$ . So, $a(t) = -6t + 20$ .
Calculate Acceleration at $t=1$ : Now, we evaluate the acceleration function $a(t)$ at $t=1$ to find the acceleration of the particle at that time. Substituting $t=1$ into $a(t)$ gives us $a(1) = -6(1) + 20$ .
Calculate Acceleration at $t=1$ : Now, we evaluate the acceleration function $a(t)$ at $t=1$ to find the acceleration of the particle at that time. Substituting $t=1$ into $a(t)$ gives us $a(1) = -6(1) + 20$ .Calculating $a(1)$ gives us $-6 + 20$ , which equals $14$ . Therefore, the acceleration of the particle at $t=1$ is $14$ meters per second squared.