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A particle moves along the
x-axis so that at time
t >= 0 its velocity is given by
v(t)=18t^(2)-72 t+54. Determine all intervals when the speed of the particle is increasing.

A particle moves along the $x$ -axis so that at time $t \geq 0$ its velocity is given by $v(t)=18 t^{2}-72 t+54$ . Determine all intervals when the speed of the particle is increasing.

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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)=18 t^{2}-72 t+54

. Determine all intervals when the speed of the particle is increasing.

Calculate Velocity Function: To determine when the speed of the particle is increasing, we need to find when the acceleration, which is the derivative of the velocity, is positive. The velocity function is given by $v(t) = 18t^2 - 72t + 54$ .
Calculate Acceleration Function: First, we calculate the derivative of the velocity function to get the acceleration function $a(t)$ . The derivative of $18t^2$ is $36t$ , the derivative of $-72t$ is $-72$ , and the derivative of a constant, $54$ , is $0$ . So, $a(t) = 36t - 72$ .
Determine Positive Acceleration: Next, we find when the acceleration is positive. This occurs when a(t) > 0. So we need to solve the inequality 36t - 72 > 0.
Solve Acceleration Inequality: To solve the inequality 36t - 72 > 0, we add $72$ to both sides to get 36t > 72.
Isolate Variable $t$ : Now, we divide both sides by $36$ to isolate $t$ , which gives us t > 2.
Conclusion: The solution to the inequality indicates that the acceleration is positive for t > 2. Therefore, the speed of the particle is increasing for all time intervals when t > 2.

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