# Example: Velocity at the top

#### Prerequisites

## Understanding the situation

The concept of instantaneous velocity is a bit of a tricky one. To get a velocity, you need to compare the position at two times, but since it is "instantaneous" we are defining it at an instant — at a single time. Of course this same problem occurs with trying to define any derivative. A derivative is about change -- which requires looking at two things — but we define the derivative at a point which is one thing. Perhaps this makes most sense geometrically. Consider the black curve at the right of position (x) as a function of time (t). If we want the velocity, we want the slope of the curve.

The average velocity between times $t_1$ and $t_2$ is the slope of the red line =(change in $x$) / (change in $t$). As those two times get closer and closer together, the red line approaches the blue line: the line that is tangent to the curve at $t$ (touches the curve at only one point). The slope of that line is the derivative of $x$ at the point $t$, the instantaneous velocity.

## Presenting a sample problem

When a small dense object it thrown straight up at not too high a speed, it's motion can be approximately described as having a constant acceleration pointing downward. Suppose such an object is thrown upward with an initial velocity $v_0$ at a starting time $t_0$.

(a) Find equations that describes its velocity and position as a function of time, assuming that the height is given by the coordinate $y$ and the positive direction is up.

(b) Find the maximum height the object reaches and determine the velocity at that instant of time.

(c) Discuss your findings.

## Solving this problem

(a) The object is thrown straight up so all we have to worry about it $y$ as a function of $t$. It doesn't tell us where the origin is. What is does tell us is that the acceleration is constant and pointing down. It tells us to take the $y$ coordinate as positive upward and let's take $t=0$ when we release the ball. Then the acceleration is a negative number. For reasons that will be discussed later, let's call this number "$-g$". This means that "$g$" is a positive number. This tells us:

$$a(t) = \frac{dv}{dt} = -g$$

(We've written $a(t)$ to remind ourselves that a is usually a function of time. The equation tells us that in this case, it is the constant function.) The only function that has a constant derivative is a linear function: $y = mx+b$ in an x-y plot. In a v-t plot it would be $v = mt+b$.

With this function, the derivative of $v$ would be $\frac{dv}{dt} = m$ so we must take $m = -g$. What's $b$? If we look at $v= -gt + b$ and set $t = 0$ we get v$ = b$. So we have to interpret $b$ as the velocity we have at time $t = 0$.

In order that the name remind us what it represents physically, let's write $b = v_0$ (the velocity at time 0). This gives our equation for $v$ as a function of time to be

$$v(t) = -gt + v_0$$

Usually we write this with the constant term first, but it doesn't make any difference (since addition is commutative).

$$v(t) = v_0 -gt $$

That's the first part of (a). Now we have to get the position, $y(t)$. Since $v = dy/dt$, we are looking for a function whose derivative is a linear function. Since we know the derivative reduces the exponent by one, we expect the answer will be a quadratic, $y(t) = at^2 + bt + c$. To see what $a$, $b$, and $c$ are let's take the derivative and set it equal to $v$.

$$\frac{dy}{dt} = \frac{d}{dt} (at^2 + bt + c) = 2at + b$$

For this to be the correct $v$, we have to have the coefficients of $t$ and the constant to match, so $2a = -g$ or $a = -½g$ and $b = v_0$

This gives the physical meaning of the mathematical constants. What about $c$? Clearly, $c = y(0)$ so it's the starting value. We don't actually know what this is (we have no reason to choose the origin of the coordinate where we happen to release the object) so let's call it $y_0$. This gives the equation for $y$

$$y(t) = -½gt^2 + v_0t + y_0$$

Typically in physics we write this in the opposite way the mathematicians do; as

$$y(t) = y_0 + v_0t -½gt^2 $$

Although it doesn't really matter, this puts where the object starts first, then the change due to its initial motion, then the change as a result of that motion changing. This arrangement mimics how we think about the physical motion.

(b) There are lots of ways to do this. But since we have the equations, let's use some math. We know from calculus that the maximum of a function occurs when it turns around — changes from increasing to decreasing — and that occurs when its derivative (the slope of the tangent to the curve) is zero (is horizontal). Let's just differentiate and solve for the time at which y reaches its maximum, t_{m}.

$$\frac{dy}{dt}|_{t_m} = v_0 - gt_m = 0$$

$$v_0 = gt_m$$

$$t_m = \frac{v_0}{g}$$

Putting this into our function gives the maximum value of $y = y_m$.

$$y_m = y_0 + v_0 t_m -½gt_m^2 = y_0 + v_0 \frac{v_0}{g} -½g(\frac{v_0}{g})^2$$_{ }

$$y_m = y_0 + \frac{ v_0^2}{g} - \frac{v_0^2}{2g}$$

$$y_m = y_0 + \frac{v_0^2}{2g} $$

Since it started at y_{0}, the height it rises is $v_0^2/2g$ (which you could have totally gotten — except for the factor of 2 — from a dimensional analysis).

(c) Since the derivative of the position is the velocity, the condition that the derivative is equal to zero (to find the maximum) is precisely the condition that the velocity is zero. This makes perfect sense since if the velocity is positive, the object is still going up it isn't at the top since in a moment it will be higher. If the velocity were negative, the object would be going down and a moment earlier it would have been higher. The top must be the place where the velocity switches from positive to negative.

But notice that the velocity is only zero instantaneously. A straight line ($mx + b$ or $v_0 - gt$) doesn't "stop" when it crosses through zero. It's only zero at one point. So although the velocity is 0 at the top, the object doesn't stop since the velocity is changing: the acceleration is not zero. Our everyday definition of "stop" is too crude to distinguish between an object being at rest (all derivatives of position = 0) and only the velocity being zero for only an instant of time — not an interval!

Joe Redish 12/31/14

Last Modified: February 24, 2019