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A particle moves along the
x-axis so that at time
t >= 0 its velocity is given by
v(t)=3t^(2)-24 t+36. Determine all intervals when the acceleration of the particle is positive.

A particle moves along the $x$ -axis so that at time $t \geq 0$ its velocity is given by $v(t)=3 t^{2}-24 t+36$ . Determine all intervals when the acceleration of the particle is positive.

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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 velocity is given by

v(t)=3 t^{2}-24 t+36

. Determine all intervals when the acceleration of the particle is positive.

Find Acceleration Function: To find the intervals when the acceleration is positive, we first need to find the acceleration function, which is the derivative of the velocity function. $\newline$ Given velocity function: $v(t) = 3t^2 - 24t + 36$
Calculate Derivative: Calculate the derivative of the velocity function to get the acceleration function. $\newline$ Acceleration, $a(t) = v'(t) = \frac{d}{dt} [3t^2 - 24t + 36]$ $\newline$ $a(t) = \frac{d}{dt} [3t^2] - \frac{d}{dt} [24t] + \frac{d}{dt} [36]$ $\newline$ $a(t) = 6t - 24$
Determine Positive Acceleration: To determine when the acceleration is positive, we set the acceleration function greater than zero and solve for $t$ .6t - 24 > 06t > 24t > 4
Solve for Positive Acceleration Interval: The acceleration is positive for all $t$ values greater than $4$ . Therefore, the interval when the acceleration is positive is $(4, \infty)$ .

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