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Velocity vs. time graphs gives us valuable information about the motion of a moving object.

The slope of velocity-time graphs gives us average and instantaneous acceleration, The area under v-t graphs represents the displacement and distance traveled, Moments when the graph intersects the time axis, indicating turning points.

In this article, we want to learn about these topics in detail, with some solved examples.

Consider a moving object so that at any instant of time you have its velocity (instant velocity). If you put these instantaneous velocities on a vertical axis, called velocity, and time on the horizontal axis, you would arrive at a so-called velocity-versus-time graph.

Therefore, a velocity-time graph is nothing except the velocity of a moving body at any instant of time whose vertical axis has SI units of m/s.

Like position vs. time graphs, such diagrams provide a snapshot of all the kinematic variables involved in the motion.

Velocity vs. Time Graphs: Slopes

Certainly, you learned in math class, the slope of a function, in the x-y plane, is defined as a change in the vertical axis over a change in the horizontal axis or in math language \[\text{slope}=\frac{\Delta y}{\Delta x}\] Replacing the y-axis with velocity (v) and x-axis with time (t), we obtain a v-t graph whose corresponding slope is defined as \[\text{slope}=\frac{\Delta v}{\Delta t}\] What formula in physics does this slope remind you of? Average acceleration.

Recall that average acceleration in physics is defined as a change in velocity divided by a change in time, $\bar{a}=\frac{\Delta v}{\Delta t}$.

Combining these two expressions, we arrive at the following important result:

The slope of a velocity-time graph gives the acceleration of a moving object.

But which type of acceleration, average or instantaneous acceleration?

To answer this question, we must pay attention to the difference between the slope of a line segment and of a tangent line.

If we select two points on the velocity vs. time graph in a finite time interval, then there is a finite difference between their corresponding velocity values.

In this case, the slope of the line segment between these two points is the average acceleration.

Now, consider the two points which you selected are so close to each other so that, in practice, they coincide with each other. In this case, finite difference $\Delta v$ converts to infinitesimal change $dv$, and similarly for time $\Delta t \to dt$.

Put it simply, here, the slope of the tangent line to the velocity-time graph gives us the instantaneous acceleration.

Keep in mind that, when the velocity vs. time graph of a moving object is a straight line, the slope of a line segment in that interval equals the slope of the tangent line at each point there.

In the next example, we see how to find slopes in a velocity-time graph.

**Example: Find (a) the slope of the line segment $AB$, $BC$, and $CD$
(b) And the slope of tangent lines at points $E$ and $Z$. **

**Solution**: Slope is defined as the ratio of the difference in the vertical axis to a change in the horizontal axis.

For the line segment $AB$, the starting point is A with coordinate $A(v=5,t=0.5)$, and ending point is $B(v=2,t=1.5)$. The change in vertical $\Delta v$, and horizontal $\Delta t$ axes are \begin{gather*}\Delta v=v_B-v_A=2-5=-3\\\Delta t=t_B-t_A=1.5-0.5=1\end{gather*}Next, construct their ratio that gives us the slope in the interval $[t=0.5,t=1.5]$ \[\text{slope}=\frac{\Delta v}{\Delta t}=\frac{-3}{1}=-3\] The slope is negative.

Similarly, the slope of segment $BC$ is \[\text{slope}=\frac{\Delta v}{\Delta t}=\frac{4-2}{2-1.5}=4\]

And also, the slope of $BC$ is zero because there is no change in the vertical axis.

(b) Since the given graph is composed of straight lines, so the slope of the tangent line equals the slope of the line segment in that interval.

Points $E$ and $Z$ lie in the segments $AB$ and $BC$, respectively, so their slopes are also 1 and -3.

There is an important note about the sign of slopes in a v-t graph. When the straight line makes an acute angle with the positive $x$ axis, the slope is positive, and an obtuse angle makes a negative slope. This is shown in the figure below.

In the next example, the average and instantaneous acceleration of a moving object is obtained using a v-t graph.

**Example (1): The velocity of a moving car along a straight line as a function of time is depicted in the following graph. Find
(a) Average acceleration in the time interval $1\,{\rm s}$ to $2\,{\rm s}$.
(b) Average acceleration during the next 2 seconds.
(c) Instantaneous acceleration at points $A$, $C$, and $D$.**

**Solution**: As mentioned, the average acceleration is the slope of the line segment joining two arbitrary points on the v-t graph whereas instantaneous acceleration is the slope of a tangent line at a specific instant of time.

(a) First, locate the corresponding values of velocities in the given time interval \begin{gather*} t=1\,{\rm s}\quad \to \quad v=1\,{\rm m/s}\\t=2\,{\rm s}\quad \to \quad v=5\,{\rm m/s}\end{gather*}

Next, by applying the definition of slope as the change in vertical axis over the change in horizontal axis, we have \begin{align*}\text{slope}&=\frac{\text{vertical change}}{\text{horizontal change}}\\\\&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{5-1}{2-1}\\\\&=4\,{\rm m/s^2}\end{align*} Therefore, during time interval $[1\,{\rm s},2\,{\rm s}]$, the car's speed changes at a constant rate of $4\,{\rm m/s^2}$.

(b) As the graph shows, there is no change in the vertical axis which is the same as the change in the car's velocity $\Delta v=0$. So, the slope in this time interval is zero or the car's velocity does not change with time.

(c) The v-t graph in this time interval is a straight line so the slope of the line segment between any two points equals the slope of the tangent line at each point in this interval.

To find the slope of the tangent line at point $A$ or the instantaneous acceleration at time $t=1.5\,{\rm s}$, select two arbitrary points in this interval and compute its slope.

One point is selected as $O$ and the other is $A$ itself. The slope of this line segment equals the slope of tangent line at point $A$. \begin{align*}\text{slope}&=\frac{\text{vertical change}}{\text{horizontal change}}\\\\&=\frac{v_A-v_O}{t_A-t_O}\\\\&=\frac{3-1}{1.5-1}\\\\&=4\,{\rm m/s^2}\end{align*} As expected, the slope at point $A$ was equal to the slope of line segment of $OB$.

Thus, as a general rule keep in mind that in a straight line velocity-time graph, average acceleration equals instantaneous acceleration.

Now, we want to find the instantaneous acceleration at point $C$. In this interval, the line segment is horizontal so $\Delta v=0$ and the slope is zero. So, at each point in the time interval of $[2\,{\rm s},4\,{\rm s}]$, the instantaneous acceleration is zero.

At point $C$, the situation is different. We don't have the value of time at this point but we see that point $C$ lies in a straight line where the coordinates of their initial and final points are known. So, the slope between these two known points is \[\text{slope}=\frac{v_f-v_i}{t_f-t_i}=\frac{2-5}{4-3.5}=-6\,{\rm m/s^2}\] Therefore, the instantaneous acceleration at point $C$ is $-6\,{\rm m/s^2}$.

One of the important information that is extracted from a velocity-time graph, is the *direction of motion*. In the next, we will find this direction by a velocity-versus-graph.

**Example (2): The velocity vs. time graph for an object moving along a straight path is shown below. Discuss the direction of the motion throughout the path. **

**Solution**: Recall that the direction of a motion is determined by the direction of its velocity. In a velocity-time graph, a negative velocity indicating a movement toward the negative $x$-axis and a positive value for velocity shows movement along the $+x$-axis.

Here, in the time interval $1\,{\rm s}$ to $2\,{\rm s}$ all corresponding values of velocity are positive so the object is moving along the positive $x$-axis.

$t=2\,{\rm s}$ is the instance when the graph enters into the negative values for velocity. Namely, at this moment the object stops and changes its direction toward the $-x$-direction. Such moments are called turning points.

During $2\,{\rm s}$ to $4\,{\rm s}$, the graph lies all the way in the negative velocity ranges. So, in this interval, the object is moving along the negative $x$-direction.

Again, at $t=4\,{\rm s}$, the object enters into the positive values of velocity. So, this is another turning point.

After that moment, the graph lies totally in the positive values of velocity. So, its motion is along the positive $x$-direction.

In this example, we found out that the *turning points of a velocity-time graph are when the graph stops $v=0$ and changes its direction*.

**Example (3): Give a physical interpretation for a moving car having the following velocity vs. time graph. **

**Solution**: Assume the car is initially moving along the positive $x$-axis.

As you can see, all graph points lie in the range of positive values of velocity, so the car retains its initial direction of motion along the positive $x$-axis.

For the first 2 seconds of motion, the car increases its speed at a constant rate (constant acceleration) because the slope in this interval does not change.

Next, at the instant of $t=2\,{\rm s}$, the car fixes its speed at 5 m/s for the next 1.5 seconds along the same previous direction.

At the moment of $t=3.5\,{\rm s}$, the car starts to slow down its velocity for the next 0.5 seconds to the final velocity of 2 m/s, still moving in the positive $x$-axis.

**Example (4): Two cars are moving along a straight path with the following velocity vs. time graph. Both of them accelerate from 0 to 100 km/h in a time interval of 10 s. Compare the following for the two cars
(a) The average acceleration.
(b) The instantaneous acceleration.**

**Solution**: The slope of a line segment between two arbitrary points in a v-t graph gives the average acceleration and the slope of a tangent line at a particular point represents the instantaneous acceleration.

(a) The average acceleration is the change in velocity over the change in time elapsed. The graph shows us (segment in green) that for both cars, these changes (vertical and horizontal) are the same, so the average accelerations of both cars are equal. \[\bar{a}_A=\bar{a}_B=\frac{100}{10}=10\,{\rm m/s^2}\]

(b) The instantaneous acceleration, which is the same as the slope of tangent lines to the v-t graph, during this time interval for two cars are shown by red lines.

As you can see, initially the tangent to the car $A$ makes a bigger angle than the car $B$, so the car $A$ starts its motion with a larger acceleration.

The angles for each tangent show a measure of instantaneous acceleration at that instant of time.

In the first 2 seconds of motion, the top curve is steeper with a larger angle (bigger instantaneous acceleration).

At the moment of $t=4\,{\rm s}$, this trend of slopes changes, and the tangent's slope of the car $B$ is greater than $A$ for the remaining last 6 seconds of motion.

Hence, after the instant of 4 s, the instantaneous acceleration of the car $B$ is greater than $A$.

In the next example, the various cases of a motion that appear in a velocity-time graph are discussed.

**Example (5): Describes each of the following velocity vs. time graphs.**

**(a) Three straight lines with positive slope.**

The red line has started its motion at some positive velocity along the positive $x$-axis, increases it at a positive constant rate (positive slope = acute angle = positive acceleration)

The blue line has the same description of motion but it starts at rest (initial speed is zero).

The green line describes the motion of an object that starts its movement at some negative velocity. This means that the object travels to the negative $x$-axis initially for $t$ seconds.

At that instant, its velocity becomes zero, turn back toward the positive $x$ axis (since after $t$, the velocities are positive)

The red line starts its motion at some positive initial velocity and increases it at a constant rate (constant slope=constant acceleration).

**(b) Two straight lines with negative slopes.**

**Solution**: The straight lines make obtuse angles with the positive $x$-axis so their accelerations are negative.

The red line describes a motion in which the object starts its movement at some initial speed along the positive $x$-axis, decreases it at a constant rate in $t$ seconds, stops at that moment, reverses its direction, and moves back toward the negative $x$-axis.

**(c) Two horizontal straight lines.**

**Solution**: The green line is for an object moving away from the origin at a constant velocity toward the positive $x$-axis.

The red line is a description of motion in which the object moving away from the origin at a constant velocity but to the negative $x$-axis.

Velocity vs. Time Graphs: Area

Until now, we learned how to find acceleration and the direction of a motion along a straight path using a velocity-time graph.

The other important information that this graph gives us is the distance traveled or the displacement of the moving object.

As a rule of thumb, the area under the velocity vs. time graph in some time interval represents the displacement in that time interval.

The proof of this rule requires some calculus that is beyond the scope of this article.

**Example (6): The velocity-time graph of a runner is shown below. How far does the runner move during the first 3 seconds?**

**Solution**: ``how far'' means the displacement of the runner. We know that the area under the velocity-versus-time graph gives the displacement during a particular time interval. Thus, the displacement is the area of the shaded triangle between $t=0$ and $t=3\,{\rm s}$.

\begin{align*} \Delta x&= \text{triangle's area}\\\\&=\frac 12 \times base \times height\\\\&=\frac 12 \times 3\times 9\\\\&=13.5\quad {\rm m}\end{align*} Thus, the runner moves 13.5 meters in 3 seconds along the straight path.

Note that the area under a velocity-time graph has the SI units of $m$.

In the next example, we answer this question about *how to find the distance in a velocity-versus-time graph.*

**Example (7): The velocity-time graph of a moving car along a straight path is shown below. In the time interval $t_1=0$ to $t_2=5\,{\rm s}$, find (a) displacement, (b) distance traveled.**

**Solution**: Area under a velocity vs. time graph gives the displacement. But, pay attention to this note that the displacement is a vector in physics having both a magnitude and a direction.

As you can see, the total area in this problem consists of two areas, a blue triangle, and a yellow trapezoid.

The areas of a triangle and a trapezoid are calculated as below \begin{gather*}\text{triangle's area's}=\frac{b\times h}{2}\\\\ \text{trapezoid's area}=\frac{a+b}{2}\times h\end{gather*} Thus, substituting the values gives \begin{gather*}\text{triangle's area}=\frac{2\times (-4)}{2}=-4 \\\\ \text{trapezoid's area}=\frac{2+3}{2}\times (2)=5\end{gather*} Note that here the areas can be a negative in contrast to math.

A negative area under a velocity-time graph shows the direction of the displacement vector.

In this case, the negative area of the triangle indicating a net displacement toward the negative $x$-axis.

Hence, the total area, which is the algebraic sum of areas, is the same as the displacement. \[\text{displacement=total area}=(-4)+5=1\quad {\rm m}\]

(b) The absolute value of area under a $v-t$ graph gives the distance traveled. \[\text{total distance}=|-4|+5=9\quad{\rm m}\]

**Example (8): The velocity-time graph of a trip is shown below. Find the total displacement of the moving object.**

**Solution**: The bounded area under a velocity vs. time graph gives the displacement. Here, the bounded area is a trapezoid that is colored.

The upper base is 2 units and lower base is 6 units. Its height is also 3 units. The area of a trapezoid is found the formula \begin{align*} A=\frac{\text{upper base+lower base}}{2}\times height\\\\=\frac{2+6}{2}\times(3)\\\\=12\quad {\rm m}\end{align*}

**Example (9): The velocity vs. time graph for a trip is shown below.
(a) Find the acceleration for each section.
(b) The total distance traveled from $5\,{\rm s}$ to $7\,{\rm s}$.**

**Solution**: The slope of a velocity vs. time graph represents the acceleration of an object. The slope is also a change in the vertical axis over a change in the horizontal axis.

(a) For segment $A$, the vertical change is $\Delta v=v_2-v_1=3-0=3\,{\rm m/s}$ and horizontal change is $\Delta t=2-1=1\,{\rm s}$. Thus, the slope or acceleration is \[a_A=\frac{\Delta v}{\Delta t}=\frac 32\quad{\rm m/s^2}\] For section $B$, we have \[a_B=\frac{v_2-v_1}{t_2-t_1}=\frac{3-3}{4-2}=0\] The slope or acceleration of segment $C$ is \[a_C=\frac{v_2-v_1}{t_2-t_1}=\frac{9-3}{5-4}=6\quad{\rm m/s^2}\] And finally, the acceleration of section $D$ is found to be \[a_D=\frac{v_2-v_1}{t_2-t_1}=\frac{11-9}{9-5}=0.5\quad {\rm m/s^2}\]

(b) Area under a velocity-time graph gives displacement vector which can be also a negative value. The absolute value of this area also gives the distance traveled in that direction.

The bounded area under this $v-t$ graph during the time interval $[5\,{\rm s},7\,{\rm s}]$, is a the colored area of a trapezoid

\begin{gather*} \text{displacement=trapezoid's area}\\\\=\frac{9+10}{2}\times (2)\\\\=19\quad {\rm m}\end{gather*} Thus, this object moves a distance of 19 meters along the direction of motion.

Note that the area of a trapezoid is found by using the formula \[A=\frac{\text{upper base+lower base}}{2}\times height\]

**Example (10): In the following velocity-time graph, find
(a) The acceleration for each section.
(b) The total displacement traveled in 7 s.
(c) The total distance traveled in 7 s.
(d) What is the displacement after 7 seconds. **

**Solution**: (a) The slope of a line on a velocity-time graph gives the acceleration of the object. Thus, the slope from 0 to 1 s is \[\text{slope}=\frac{\Delta v}{\Delta t}=\frac{6-0}{1-0}=6\,{\rm m/s^2}\] For 1 s to 3 s, the slope or acceleration is \[\text{slope}=\frac{v_f-v_i}{t_f-t_f}=\frac{-6-6}{3-1}=-6\,{\rm m/s^2}\] Where subscripts $i$ and $f$ denote the initial and final points on the line segment. And finally, for 3 s to 7 s, the acceleration is \[\text{slope}=\frac{v_f-v_i}{t_f-t_i}=\frac{0-(-6)}{7-3}=+1.5\,{\rm m/s^2}\]

(b) The bounded area under the velocity vs. time graph represents the displacement. This area can be negative since the displacement is a vector in physics and the sign of the area shows us the direction of displacement.

In this graph, in the time interval 0 to 7 s, there are two bounded areas that have been colored.

The area triangles are found as \begin{align*}\text{yellow area}&=(1/2)(base\times height)\\&=(1/2)(2\times 6)\\&=6\quad {\rm m}\\\\\text{pink area}&=(1/2)(5\times (-6))\\&=-15\quad {\rm m}\end{align*} The algebraic sum of areas under a $v-t$ graph, gives the total displacement during that time interval. So, \[\text{total area}=6+(-15)=-9\quad {\rm m}\] Thus, the total displacement of the object is 9 meters toward the negative $x$-axis.

(c) The absolute value of area under a $v-t$ graph during any time interval gives the distance traveled over that interval. In this case, the object has traveled a distance of 6 meters for 2 s, and 15 meters for the next 5 seconds. So the total distance traveled in 7 s is the sum of these distances \[\text{total distance}=6+15=21\quad {\rm m}\]

(d) As the $v-t$ plot shows, after the instant of 7 s, the velocity is zero as time goes. In other words, after this moment, the object is standing at rest. So, its displacement is zero.

In this long article, we learned what means by slopes to and areas under a velocity vs. time graph or how to find acceleration using slopes to and displacements by areas under a velocity vs. time graph.

As conclusion:

The slope of a line on a velocity-time graph gives us a geometric interpretation of the acceleration.

The bounded area under a velocity-time graph gives the displacement.

**Author**: Ali Nemati

**Page Published:** 8-18-2021

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