Suppose an object moves along the x-axis so that its position at time $t$ is $x=-t+\frac{t^3}{6}$.
(a) Find the velocity, $v(t)=x(t)$, of the object.
(b) What is $v(0)$? What does this say about the direction of motion of the object at time $t=0$?
(c) When is the object at the origin? What is the velocity of the object when it is at the origin?
x(t) = -t + t^3/6
v(t) = dx/dt = -1 + t^2/2
I think you can now answer the questions if you think a bit.
To find the velocity $v(t)$ of the object as a function of time $t$, we need to take the derivative of the position function $x(t)$. The velocity is defined as the rate of change of position with respect to time.
(a) To find $v(t)$, we take the derivative of $x(t)$ with respect to $t$. Using the power rule of differentiation, we differentiate each term separately:
$\frac{dx}{dt} = \frac{d}{dt}(-t+\frac{t^3}{6})$
$\frac{dx}{dt} = -1 + \frac{1}{6} \cdot 3t^2$
$\frac{dx}{dt} = -1 + \frac{t^2}{2}$
So, the velocity function $v(t)$ is $v(t) = -1 + \frac{t^2}{2}$.
(b) To find $v(0)$, we substitute $t=0$ into the velocity function:
$v(0) = -1 + \frac{0^2}{2}$
$v(0) = -1$
The velocity at $t=0$ is $-1$. This means that the object is moving in the negative direction (to the left) at $t=0$.
(c) To find when the object is at the origin, we set $x(t) = 0$ and solve for $t$. The origin corresponds to when the object is at position $x=0$.
$x(t) = -t + \frac{t^3}{6} = 0$
To solve this equation, we can factor out $t$:
$t\left(-1 + \frac{t^2}{6}\right) = 0$
This equation is satisfied when $t=0$, giving us one solution. For the other solution, we solve the equation:
$-1 + \frac{t^2}{6} = 0$
$\frac{t^2}{6} = 1$
$t^2 = 6$
$t = \pm \sqrt{6}$
So, the object is at the origin at $t=0$ and $t=\sqrt{6}$. To find the velocity of the object at the origin, we substitute $t=0$ into the velocity function:
$v(0) = -1 + \frac{0^2}{2}$
$v(0) = -1$
The velocity of the object at the origin is $-1$.