Question

X(t)=t^2-2t+1 when does it comes to rest ? andwhat does it mean∆v=0&∆v=constant​

Answer 1

Hello!

Recall that x(t) is an object's displacement over time.

For an object to be at rest, its instantaneous VELOCITY must equal 0 m/s, or there is no change in the object's displacement at that time.

The object's velocity over time, v(t), is simply the derivative of the position function.

We can use power rules to find an expression for the object's velocity:
$\frac{dy}{dx} x^n = nx^{n-1}$

Differentiate:
$x(t) = t^2 - 2t + 1\\\\x'(t) = v(t) = 2t - 2$

Now, since v(t) must equal 0 m/s for the object to be at rest, we can solve for 't' by setting the equation equal to 0.

$0 = 2t - 2\\\\2 = 2t\\\\\boxed{t = 1s}$

Now, assuming there's another part to your question:

Δv = 0 indicates that there is NO CHANGE in the object's velocity, which means that the object is not experiencing a net acceleration. Therefore, a = 0 m/s², or the graph of v(t) would be a straight, horizontal line.

Δv = constant indicates that there is a CONSTANT change in the object's velocity caused by a constant, non-zero acceleration. The graph of a(t) would be a straight, horizontal line, and the graph of v(t) would be linear and sloping.