If the position of a particle is $x(t)=t^2-4t+5$, at what time is the particle at rest?

Key takeaway: A particle is at rest when its velocity is zero. Since velocity is the derivative of position, the task is to differentiate $x(t)$ and solve for $t$.

The particle is at rest when its velocity equals zero.

Given the position function

$x(t)=t^2-4t+5,$

the velocity is the derivative:

$v(t)=x'(t)=2t-4.$

Set velocity equal to zero:

$2t-4=0.$

Solve:

$2t=4,$

$t=2.$

Since the question asks for $t\ge 0$, the answer is

$\boxed{t=2}.$

---

Pitfalls the pros know 👇 Do not confuse the position function with the velocity function. The particle is not at rest when $x(t)=0$; it is at rest when $x'(t)=0$.

What if the problem changes? If the position function were changed to something like $x(t)=at^2+bt+c$, the same method would apply: differentiate to get $v(t)=2at+b$, then solve $2at+b=0$ for the rest time.

Tags: position function, velocity, derivative