A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=-t^(2)-8t+44. Determine the acceleration of the particle at t=9. Answer:

Find Derivative of x(t): 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, let's find the first derivative of x(t) which is the velocity function v(t). The derivative of x(t)=-t^(2)-8t+44 with respect to t is v(t) = dx/dt = -2t - 8.Find Acceleration Function: Now, let's find the second derivative of x(t) which is the acceleration function a(t). The derivative of v(t) = -2t - 8 with respect to t is a(t) = dv/dt = -2.Acceleration is Constant: Since the second derivative is a constant, the acceleration of the particle does not depend on time t. Therefore, the acceleration of the particle at t=9 is the same as at any other time.Calculate Acceleration at t=9: The acceleration of the particle at t=9 is a(t) = -2.

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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^(2)-8t+44. Determine the acceleration of the particle at
t=9.
Answer:

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=-t^{2}-8 t+44$ . Determine the acceleration of the particle at $t=9$ . $\newline$ 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^{2}-8 t+44

. Determine the acceleration of the particle at

t=9

\newline

Answer:

Find Derivative of $x(t)$ : 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, let's find the first derivative of $x(t)$ which is the velocity function $v(t)$ . The derivative of $x(t) = -t^{2} - 8t + 44$ with respect to $t$ is $v(t) = \frac{dx}{dt} = -2t - 8$ .
Find Acceleration Function: Now, let's find the second derivative of $x(t)$ which is the acceleration function $a(t)$ . The derivative of $v(t) = -2t - 8$ with respect to $t$ is $a(t) = \frac{dv}{dt} = -2$ .
Acceleration is Constant: Since the second derivative is a constant, the acceleration of the particle does not depend on time $t$ . Therefore, the acceleration of the particle at $t=9$ is the same as at any other time.
Calculate Acceleration at $t=9$ : The acceleration of the particle at $t=9$ is $a(t) = -2$ .