Kinematic quantities

Kinematics describes the motion of objects, without considering the forces that affect the motion. To participate in kinematics, we need to define, measure, and interpret certain quantities. Let's explore these quantities through the example of a sprinter training for the $100,\text{m}$ dash.

##Position

The sprinter wants to analyze his motion down a track. So, he defines a one-dimensional coordinate system for the track with the origin $(0)$ at the starting line, as shown below.

[[☃ image 17]]

The sprinter places a [[☃ definition 1]] at the starting line to record his position at different times. An object's position is its location relative to a chosen origin. Position is a vector quantity with both magnitude and direction. For example, if the sprinter is located $5,\text{m}$ from the starting line in the $+x$ direction, his position vector at that instant is modeled below.

[[☃ image 19]]

The sprinter sets the detector to record his position every $1,\text{s}$ for $20,\text{s}$ after he begins running. He begins at rest at the starting line, and then sprints forward in a straight line in the $+x$ direction. The motion detector records the position-time data shown on the graph below.

[[☃ image 3]]

[[☃ graded-group-set 1]]

The data collected by the motion detector represents the sprinter's position as a function of time. In other words, we can ask about a specific time as an input, and get the runner’s position at that time as an output. Mathematically we represent this as $x(t),$ pronounced $\text{``}x$ of $t\text{"}.$

If we can find a mathematical pattern that fits the data points well, we can also model the sprinter's position-time relationship using an equation. The sprinter analyzes his data and finds that there are four distinct patterns over the $20\text{-s}$ interval. He uses a computer to find a piecewise function that fits his data. The piecewise function is shown below, both in graph and equation form.

[[☃ image 5]]

$x(t) =
\begin{cases}
\color{#2e6ead}{1.50t^2} & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{11.97t - 23.94} & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.05t^3 - 2.60t^2 + 43.71t - 137.71} & \text{if } 8 < t \leq 16 \
\color{#368549}{106.50} & \text{if } 16 < t \leq 20
\end{cases}$

[[☃ explanation 1]]

[[☃ graded-group-set 2]]

The piecewise function is a model of the real motion—it's not a perfect representation, but it's a good fit. The model is useful for describing the sprinter's motion in a simplified way, and for estimating values between the collected data points. The equation also allows us to calculate other quantities for the motion.

##Velocity

Through most of the $20\text{-s}$ interval analyzed above, the sprinter's position was changing. A change in position is called displacement, $\Delta x.$ Since position is a vector, displacement is also a vector.

An object's displacement over a time interval of duration $\red{\Delta t}$ is equal to its position at the end of that interval, $x(t+\red{\Delta t}),$ minus its position at the start of that interval, $x(t):$

$\Delta x=x(t+\red{\Delta t})-x(t)$

The sprinter wants to know the rate at which his position changed throughout his run. In other words, he wants to know about his velocity. An object's average velocity, $v_\text{avg},$ for any interval is its displacement $\Delta x$ during that interval, divided by the amount of time $\red{\Delta t}$ over which the displacement occurred:

$\boxed{v_\text{avg}=\dfrac{x(t+\red{\Delta t})-x(t)}{\red{\Delta t}}=\dfrac{\Delta x}{\red{\Delta t}}}$

Average velocity is a difference quotient—it's a difference (change) in position, divided by a difference (change) in time. So on a position-time graph, the $v_\text{avg}$ of an interval is the slope of a straight line connecting the endpoints of that interval.

[[☃ graded-group-set 1]]

The three average velocities calculated above do not equal each other. This is because the sprinter's velocity was changing continuously during these intervals. The animated graph below models how $v_\text{avg}$—represented by the slope of the black line—changes as $\red{\Delta t}$ changes. As we saw in the data table above, $v_\text{avg}$ increases—the slope gets steeper—as the interval shrinks around $t=11,\text{s}.$

[[☃ image 3]]

We're closing in on the true value of the sprinter's velocity at the instant $t=11,\text{s}.$ How small can we shrink $\red{\Delta t}?$ With the power of calculus, we can shrink the time interval so it approaches zero. When we take the limit of average velocity as $\red{\Delta t \to 0},$ we get the instantaneous velocity, $v(t),$ at time $t.$ We can represent this mathematically as a limit $(\lim)$ of our average velocity difference quotient from earlier:

$\boxed{v(t)=\displaystyle\lim_{\red{\Delta t \to 0}}\dfrac{x(t+\red{\Delta t})-x(t)}{\red{\Delta t}}=\dfrac{dx}{dt}}$

This limit of a difference quotient is called a derivative. We say that instantaneous velocity is the derivative of position with respect to time. Notice that instead of $\Delta$ symbols, the derivative contains $d$ symbols. This is because the derivative represents the ratio of very small changes in $x$ and $t$.

Graphically, we no longer have the slope of a line connecting two separate points. Instead, we've shrunk the interval such that we've found the slope at a single point. The value of a function's derivative at a point is the slope of the line tangent to the curve at that point. This is modeled by the slope of the animated black line below. Notice how the slope of the tangent line—the instantaneous velocity—changes for different $t$ values.

[[☃ image 4]]

Whether we graph the function or not, the value of a function's derivative more fundamentally represents the instantaneous rate of change of the function. We can also find the derivative analytically, using the function in equation form. We saw that the runner's position as a function of time can be modeled using the following piecewise function.

$x(t) =
\begin{cases}
\color{#2e6ead}{1.50t^2} & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{11.97t - 23.94} & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.05t^3 - 2.60t^2 + 43.71t - 137.71} & \text{if } 8 < t \leq 16 \
\color{#368549}{106.50} & \text{if } 16 < t \leq 20
\end{cases}$

Using derivative rules, we can differentiate $x(t)$ to obtain an equation for $v(t).$ [[☃ explanation 3]]

$v(t) =
\begin{cases}
\color{#2e6ead}{3.00t} \qquad\qquad;;; & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{11.97} \qquad\qquad;;; & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.15t^2 - 5.20t + 43.71} \qquad\qquad;;; & \text{if } 8 < t \leq 16 \
\color{#368549}{0} \qquad\qquad;;; & \text{if } 16 < t \leq 20
\end{cases}$

The sprinter wants to visualize his equation for $v(t),$ so he graphs it and gets the velocity-time graph below. (Notice that the horizontal axis represents the same $t$ values as on the position-time graph above, but the vertical axis has changed. Instead of position in meters, it now represents velocity in meters per second.)

[[☃ image 1]]

[[☃ graded-group-set 3]]

##Acceleration

Just as the sprinter wanted to know the rate at which his position changed over time, he also wants to know the rate at which his velocity changed over time. In other words, he wants to know about his acceleration.

An object's change in velocity over a time interval of duration $\red{\Delta t}$ is equal to its velocity at the end of that interval, $v(t+\red{\Delta t}),$ minus its velocity at the start of that interval, $v(t):$

$\Delta v=v(t+\red{\Delta t})-v(t)$

[[☃ explanation 2]]

An object's average acceleration, $a_\text{avg}$, over any interval is its change in velocity $\Delta v$ over that interval, divided by the amount of time $\Delta t$ over which the change in velocity occurred:

$\boxed{a_\text{avg}=\dfrac{v(t+\red{\Delta t})-v(t)}{\red{\Delta t}}=\dfrac{\Delta v}{\red{\Delta t}}}$

On a velocity-time graph, the $a_\text{avg}$ of an interval is the slope of a straight line connecting the endpoints of that interval. Like we did previously, we can shrink the time interval so it approaches zero. When we take the limit of average acceleration as $\red{\Delta t \to 0},$ we get the instantaneous acceleration, $a(t),$ at time $t.$ We can represent this mathematically as a limit of the average acceleration difference quotient:

$\boxed{a(t)=\displaystyle\lim_{\red{\Delta t \to 0}}\dfrac{v(t+\red{\Delta t})-v(t)}{\red{\Delta t}}=\dfrac{dv}{dt}}$

Instantaneous acceleration is the derivative of velocity with respect to time. Since $v(t)$ is already the first derivative of $x(t),$ we can also say that $a(t)$ is the second derivative of $x(t).$

On a velocity-time graph, an object's instantaneous acceleration represents the slope of the line tangent to the $v(t)$ curve. More fundamentally, $a(t)$ represents the instantaneous rate of change of $v(t)$ at the time $t.$

We can also find the derivative analytically, using the functions in equation form. We saw earlier that the runner's position and velocity as functions of time can be modeled using the following piecewise equations:

$x(t) =
\begin{cases}
\color{#2e6ead}{1.50t^2} & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{11.97t - 23.94} & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.05t^3 - 2.60t^2 + 43.71t - 137.71} & \text{if } 8 < t \leq 16 \
\color{#368549}{106.50} & \text{if } 16 < t \leq 20
\end{cases}$

$v(t) =
\begin{cases}
\color{#2e6ead}{3.00t} \qquad\qquad;;; & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{11.97} \qquad\qquad;;; & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.15t^2 - 5.20t + 43.71} \qquad\qquad;;; & \text{if } 8 < t \leq 16 \
\color{#368549}{0} \qquad\qquad;;; & \text{if } 16 < t \leq 20
\end{cases}$

Using derivative rules, we can differentiate $v(t)$ to obtain an equation for $a(t).$

$a(t) =
\begin{cases}
\color{#2e6ead}{3.00} \qquad\qquad\qquad\qquad\quad & \text{if } 0 \leq t \leq 4 \
\color{#ff7b1c}{0} \qquad\qquad\qquad\qquad\quad & \text{if } 4 < t \leq 8 \
\color{#5f45a5}{0.30t - 5.20} \qquad\qquad\qquad\qquad\quad & \text{if } 8 < t \leq 16 \
\color{#368549}{0} \qquad\qquad\qquad\quad & \text{if } 16 < t \leq 20
\end{cases}$

The sprinter graphs $a(t)$ to create an acceleration-time graph. Then he stacks his graphs of $x(t), v(t),$ and $a(t)$ and analyzes them together, focusing on each time interval separately. (Notice that the horizontal axis represents the same $t$ values on all three graphs, but the vertical axes represent different quantities and units.)

[[☃ image 4]]

[[☃ graded-group-set 1]]

The functions for $x(t),,v(t),$ and $a(t)$ tell the same story of the sprinter's motion:

• From $0,\text{s}$ to $4,\text{s},$ his position increased in the $+x$ direction, at an increasing rate. He was speeding up. He was undergoing constant acceleration in the $+x$ direction.

• From $4,\text{s}$ to $8,\text{s},$ his position continued increasing in the $+x$ direction, but now at a constant rate. He was was running with constant velocity. He maintained his maximum velocity during this interval and had zero acceleration.

• From $8,\text{s}$ to $16,\text{s},$ his position continued increasing in the $+x$ direction, but now at a decreasing rate. He was slowing down. His velocity was in the $+x$ direction, but since its magnitude was decreasing, his acceleration was in the opposite direction, which in this case is the $-x$ direction. The magnitude of his acceleration changed over the interval.

• From $16,\text{s}$ to $20,\text{s},$ his position remained constant at approximately $106,\text{m}.$ He was at rest. He had zero velocity or acceleration during this interval.

The sprinter could use these models to guide his training. For example, he may want to accelerate over a longer time at the beginning of the race, maintain his peak velocity over a longer time, or make other changes to optimize his sprint.

[[☃ explanation 3]]

##Key points

• An object's position is its location relative to a chosen origin. A change in position is called displacement.

• An object's average velocity over an interval is its displacement during that interval, divided by the amount of time over which the displacement occurred. On a position-time graph, the average velocity of an interval is the slope of a straight line connecting the endpoints of that interval.

$\boxed{v_\text{avg}=\dfrac{\Delta x}{\Delta t}}$

• An object's instantaneous velocity is the derivative of its position with respect to time. On a position-time graph, the instantaneous velocity at some time is the slope of the line tangent to the curve at that point.

$\boxed{v(t)=\dfrac{dx}{dt}}$

• An object's average acceleration over an interval is its change in velocity over that interval, divided by the amount of time over which the change in velocity occurred. On a velocity-time graph, the average velocity of an interval is the slope of a straight line connecting the endpoints of that interval.

$\boxed{a_\text{avg}=\dfrac{\Delta v}{\Delta t}}$

• An object's instantaneous acceleration is the derivative of its velocity with respect to time. On a velocity-time graph, the instantaneous acceleration at some time is the slope of the line tangent to the curve at that point.

$\boxed{a(t)=\dfrac{dv}{dt}}$

• Position, velocity, and acceleration are vector quantities with both magnitude and direction.

[[☃ explanation 2]]

The rate of change of an object's position is its velocity. The rate of change of an object's velocity is its acceleration. So, starting from the object's position as a function of time, $x(t),$ we can differentiate to find its velocity as a function of time, $v(t),$ and differentiate again to find its acceleration as a function of time, $a(t).$

What if we wanted to translate between these kinematic quantities in the reverse direction? That is, can we start from $a(t),$ use it to find $v(t),$ and use that to find $x(t)?$ Instead of finding the rate of change of a quantity, we'd be starting with the rate of change, and working backwards to sum up how much change has occurred. Let's explore how to do this through the example of an elevator.

[[☃ explanation 2]]

##Measuring $a(t)$

An engineer steps into an elevator car on the fourth floor of a building. She wants to analyze the car's vertical motion. So, she defines a one-dimensional coordinate system with the origin at the fourth floor, as modeled below. She also attaches an [[☃ definition 2]] to the car's floor.

[[☃ image 22]]

The car begins at rest. When the car starts moving, the engineer begins collecting data on the accelerometer. It records the car's acceleration every $0.2,\text{s}$ until she stops collecting data. The engineer processes her data to reveal the acceleration-time data shown on the graph below.

[[☃ explanation 2]]

[[☃ image 2]]

The engineer identifies a piecewise function which models her data well. Her model for the car's acceleration as a function of time, $a(t),$ is shown below in graph and equation form.

[[☃ image 12]]

$a(t) =
\begin{cases}
\color{#368549}{-1.50} \qquad\qquad\qquad\qquad\quad & \text{if } 0 \leq t \leq 1.5 \
\color{#5f45a5}{0} \qquad\qquad\qquad\qquad\quad & \text{if } 1.5 < t \leq 3.5 \
\color{#ff7b1c}{0.60} \qquad\qquad\qquad\qquad\quad & \text{if } 3.5 < t \leq 7.3 \
\color{#2e6ead}{0} \qquad\qquad\qquad\quad & \text{if } 7.3 < t < 11.8 \
\color{#c64440}{-0.33(t-11.8)+1.31} \qquad\quad & \text{if } 11.8 \leq t \leq 19.8
\end{cases}$

[[☃ explanation 3]]

[[☃ graded-group-set 1]]

##Using $a(t)$ to find $v(t)$

The engineer wants to use her model for the car's acceleration as a function of time to determine the car's velocity at different times. The car's average acceleration over an interval is related to its change in velocity over that interval by:

$a_\text{avg}=\dfrac{\Delta v}{\Delta t}$

If acceleration is constant during an interval, then $a_\text{avg}$ for the interval equals that constant. In other words, $a_\text{avg}$ accurately represents how the object was accelerating at all times throughout the interval. So in this case, we can use $a_\text{avg}$ to determine how much the object's velocity changed during the interval. Rearranging the equation above gives:

$\Delta v = a_\text{avg}\Delta t$

We can add this change in velocity to the object's initial velocity $(v_0)$ at the start of the interval—what its velocity already was before the change occurred—to find its final velocity $(v)$ at the end of the interval:

$\begin{align}&v = v_0+\Delta v
\\
&\boxed{v = v_0 + a_\text{avg}\Delta t}\end{align}$

For example, consider the instant $t=0.8,\text{s}.$ $a(t)$ tells us the car accelerated at a constant $-1.50,\text{m/s}^2$ from $t=0,\text{s}$ until this time. We also know the car began at rest, so the initial velocity was $0,\text{m/s}$ for this interval. So, the car's velocity at $t=0.8,\text{s}$ was:

$\begin{align}&v(0.8) = v(0) + (-1.50,\text{m/s}^2)(0.8,\text{s}-0,\text{s})
\\
&v(0.8) = 0,\text{m/s} - 1.20,\text{m/s}
\\
&v(0.8) = -1.20,\text{m/s}\end{align}$

Based on the coordinate system defined by the engineer, this means the car was moving with a speed of $1.20,\text{m/s},$ downward, at $t=0.8,\text{s}.$ We can use a similar procedure to find the car's velocity at other instants.

[[☃ graded-group-set 1]]

We could do a special velocity calculation like this for any particular time. But, it would be even more useful to represent velocity as a function of time, $v(t).$ We can find $v(t)$ using the same procedure we used above to calculate $v$ at different times, but keep the time general with $t.$ So, from the piecewise function for $a(t),$ we can sum changes in velocity to get a piecewise function for $v(t):$

$v(t) =
\begin{cases}
\color{#368549}{-1.50t} \qquad\qquad\qquad\qquad & \text{if } 0 \leq t \leq 1.5 \
\color{#5f45a5}{-2.25} \qquad\qquad\qquad\qquad & \text{if } 1.5 < t \leq 3.5 \
\color{#ff7b1c}{0.60(t-3.5)-2.25} \qquad\qquad: & \text{if } 3.5 < t \leq 7.3 \
\color{#2e6ead}{0} \qquad\qquad\quad & \text{if } 7.3 < t < 11.8 \
\color{#c64440}{?} \qquad\qquad\quad & \text{if } 11.8 \leq t \leq 19.8
\end{cases}$

Now we have a function that lets us input a time value, and get the car's instantaneous velocity as an output. However, notice that the last piece of $v(t)$ is missing. This is because the car's acceleration changed throughout this interval. We know this because the acceleration has a $t$ dependence during this interval: $a(t)=\color{#c64440}{-0.33\left(\boxed{t}-11.8\right)+1.31}.$

Unlike the other intervals, the $a_\text{avg}$ for this interval does NOT equal the car's acceleration at all times during the interval. So, how can we sum the car's change in velocity throughout this interval?

So far, we've been summing velocity changes numerically. Let's view this same process graphically. For intervals with constant acceleration, we've been multiplying the $a$ value—the "height" of that portion of the $a(t)$ graph—by the amount of time $\Delta t$ that the acceleration applies—the "width" of that portion of the $a(t)$ graph. So, change in velocity during an interval is equal to the area bounded by $a(t)$ and the $t$ axis.

The accumulating change in velocity for the car is modeled as shaded area on the animated $a(t)$ graph below. Observe how the shaded area and $v(t)$ change as $t$ changes.

[[☃ image 3]]

The bounded area over an interval represents the car's change in velocity $\Delta v$ during that interval. Note that it's a signed area—area bounded above the horizontal axis represents a positive $\Delta v,$ and area bounded below the axis represents a negative $\Delta v.$

[[☃ graded-group-set 2]]

A geometric approach allows us to find the area bounded by the final $a(t)$ interval, so we can fill in the final piece of $v(t).$ The complete $v(t)$ function is shown below in equation and graph form.

$\small{v(t) =
\begin{cases}
\color{#368549}{-1.50t} & \text{if } 0 \leq t \leq 1.5 \
\color{#5f45a5}{-2.25} & \text{if } 1.5 < t \leq 3.5 \
\color{#ff7b1c}{0.60(t-3.5)-2.25} & \text{if } 3.5 < t \leq 7.3 \
\color{#2e6ead}{0} & \text{if } 7.3 < t < 11.8 \
\color{#c64440}{-0.16(t-11.8)^2+1.31(t-11.8)} \quad & \text{if } 11.8 \leq t \leq 19.8
\end{cases}}$

[[☃ image 2]]

Since the car began at rest at $t=0,\text{s},$ the function $v(t)$ represents how much $\Delta v$ has accumulated—the signed area bounded on the $a(t)$ graph—up to time $t.$ We can visualize this relationship by stacking the graphs and seeing how the value of $v(t)$ changes as the total signed area on $a(t)$ accumulates. The animation below shows the shapes of $v(t)$ (top) and $a(t)$ (bottom). The two graphs do not share a common vertical axis, but do share a common horizontal (time) axis.

[[☃ image 6]]

Both graphs tell the same story of the elevator car's motion, from two different perspectives.

[[☃ graded-group-set 3]]

##Using $v(t)$ to find $y(t)$

Acceleration is the rate of change of velocity. So, we were able to work backwards from $a(t)$ to find $v(t)$ by summing changes in velocity up to a time $t.$ These changes in velocity are represented by the signed area bounded on the graph of $a(t).$

Analogously, velocity is the rate of change of position. So in the same way, we can work backwards from $v(t)$ to find $y(t)$—the car's position as a function of time—by summing changes in position up to a time $t.$ Recall that a change in position is also called a displacement. So, changes in position, called displacements, are represented by the signed area bounded on the graph of $v(t).$

[[☃ graded-group-set 1]]

For each of the first four pieces of $v(t),$ we can find the car's displacement with a single area formula. But the final piece of $v(t)$ poses a new challenge with its curved side. There is no simple area formula for this curved shape, so how can we calculate the displacement during this interval?

The key is to divide the interval into smaller subintervals, each with width $\red{\Delta t}.$ We represent the area bounded during each subinterval using a rectangle that extends to the curve. By summing the areas of all these rectangles, we can approximate the area bounded by the curved region. The area approximation becomes more accurate as we use more rectangles, each with a smaller width $\red{\Delta t},$ as modeled in the animation below.

[[☃ image 6]]

Each rectangle's area is a product: it's the rectangle's height (the velocity at the time of that rectangle, which we'll call $v(t_i)),$ multiplied by the rectangle's width $(\red{\Delta t}).$ The sum $\left(\sum\right)$ of these products gives us an estimate of the area bounded by the curve over the interval, which represents the car's displacement $\Delta y$ during the interval:

$\Delta y \approx\sum{v(t_i)\red{\Delta t}}$

Calculating the areas of all these rectangles by hand would be tedious. But a computer can do it quickly, so in practice we can make $\Delta t$ very small so the area estimate becomes very accurate. If we take the limit $(\text{lim})$ of this sum as the width of each subinterval approaches zero $(\red{\Delta t \to 0})$ and number of rectangles grows to infinity, our area estimate becomes perfect. This limit is called a definite integral:

$\boxed{\Delta y=\displaystyle\lim_{\red{\Delta t \to 0}}\sum{v(t_i)\red{\Delta t}}=\int_{t_1}^{t_2}v(t),dt}$

In the integral, the $\sum$ symbol is replaced by $\int,$ a stretched S that stands for sum. And $\Delta t$ is replaced by $dt,$ representing an infinitesimally small change in time. These new symbols together represent that we're summing an infinite number of infinitesimally small areas. $t_1$ and $t_2$ are the bounds of integration. They represent the start and end times of the interval we want to sum over.

[[☃ explanation 2]]

Since the car began at the origin $(y(0)=0),$ the function $y(t)$ represents how much $\Delta y$ has accumulated up to time $t.$ The engineer sums the signed areas bounded on the graph of $v(t)$ to find $y(t).$ Then, she stacks her graphs of $y(t),$ $v(t),$ and $a(t)$ and analyzes them together.

[[☃ explanation 3]]

[[☃ image 5]]

[[☃ explanation 1]]

[[☃ graded-group-set 2]]

The functions for $y(t),,v(t),$ and $a(t)$ tell the same story of the elevator's motion:

• From $0,\text{s}$ to $1.5,\text{s},$ it moved in the $-y$ direction, at an increasing rate. It was moving down and speeding up, with constant acceleration in the $-y$ direction.

• From $1.5,\text{s}$ to $3.5,\text{s},$ it continued moving in the $-y$ direction, but now at a constant rate. It was moving down at constant speed, with zero acceleration.

• From $3.5,\text{s}$ to $7.3,\text{s},$ it continued moving in the $-y$ direction, but now at a decreasing rate. It was moving down and slowing down, with constant acceleration in the $+y$ direction.

• From $7.3,\text{s}$ to $11.8,\text{s},$ its position remained constant. It was at rest. It had zero velocity or acceleration during this interval.

• From $11.8,\text{s}$ to $19.8,\text{s},$ it moved in the $+y$ direction. It was moving up, first speeding up and then slowing down, with changing acceleration.

##Key points

• An object's change in position (displacement) during an interval is equal to the signed area bounded on a graph of the object’s velocity as a function of time, $v(t).$

• If the graph of $v(t)$ is straight during an interval, it forms a polygon with the $t$-axis, and the bounded area can be found using geometry. But if the graph of $v(t)$ is curved during an interval, the bounded area must be found by dividing the interval into infinitesimally small subintervals and summing their areas. The bounded area is equal to a definite integral:

$\boxed{\Delta y=\displaystyle{\int_{t_1}^{t_2}v(t),dt}}$

• The total bounded area up to time $t$ must be added to the object's initial position at $t=0,$ to give its position at time $t.$

$\boxed{y(t)=y(0)+\displaystyle{\int_{0}^{t}v(\tau),d\tau}}$

In the integral above, the $t$ inside the integral has been replaced by $\tau$. This is simply to avoid confusion with $t$ as the upper bound of integration. The $\tau$ is a "dummy variable"—it has no effect on the result of the integral. Regardless of what we call the dummy variable, the integral says that we're summing area under the velocity function using infinitesimally small rectangles.

• An object's change in velocity during an interval is equal to the signed area bounded on a graph of the object’s acceleration as a function of time, $a(t).$

• Again, it may be possible to find the bounded area on $a(t)$ using geometry. But if the graph of $a(t)$ is curved, the bounded area must be found by dividing the interval into infinitesimally small subintervals and summing their areas. The bounded area is equal to a definite integral:

$\boxed{\Delta v=\displaystyle{\int_{t_1}^{t_2}a(t),dt}}$

• The total bounded area up to time $t$ must be added to the object's initial velocity at $t=0,$ to give its velocity at time $t:$

$\boxed{v(t)=v(0)+\displaystyle{\int_{0}^{t}a(\tau),d\tau}}$

[[☃ explanation 2]]