Blog & News
kinematic

# Acceleration on Position-Time Graph

Acceleration in a position vs. time graph can be obtained by the given initial position and velocity of a moving object.

In this article, we want to show you how to find constant acceleration using a position-time graph with some solved problems.

## Types of Motion:

An object can move at a constant speed or have a changing velocity.

Suppose you are driving a car at a constant speed of 100 km/h along a straight line. What does mean by 100 km/h?

The speed of 100 km/h indicates that you drive the first 100 km in the first hour, the next 100 km during the second hour, another 100 km for the third hour, and so on.

Now, once we plot these successive displacements as a function of time, we arrive at a straight line.

Now, when the car has a changing velocity, and plot the positions of each point the car passed through it, you arrive at an arbitrary curve, in contrast to the previous case.

In the first motion, during each hour, displacements are equal. This type of motion is called uniform motion. Because, by definition of average acceleration, the object's acceleration is zero.

But, the second case has a changing velocity, so the acceleration of the motion is not zero.

Therefore, a motion in which equal displacements occur during any successive equal-time interval is known as uniform motion.

For a constant velocity motion (uniform), the average and instantaneous velocity are always equal at all times.

According to the definition of average velocity, $\bar{v}=\frac{x-x_0}{\Delta t}$, we can find the position of a moving object at any time by rearranging it as below $x=vt+x_0$ This is a linear equation with the following position vs. time graph.

For an accelerated motion, the position of a moving object at any time is given by the kinematics equation, $x=\frac 12 at^2 +v_0t+x_0$.

This equation tells us that for an accelerated motion, position varies with time in a quadratic form whose graph is as below

Overall, a constant velocity (uniform) motion has a straight line position-versus-time graph, but a curve in the $x-t$ graph represents an accelerated motion.

This curve for a constant acceleration has a simple form of quadratic.

## Direction of acceleration on a position-time graph

Acceleration is a vector quantity in physics with both a direction and a magnitude. We can find both using a $x$-$t$ graph.

As mentioned above, the position of a uniform accelerating object varies with time as a quadratic function.

A positive acceleration yield a position-time graph opening upward.

Similarly, a negative acceleration produces a position-time graph opening downward. Both cases are shown in the figure below

## Magnitude of acceleration on a position-time graph

It is a difficult task to find the magnitude of the acceleration vector for a moving object from its position-time graph.

If the object increases its speed at a constant rate, then its acceleration is a constant value during that time interval.

In such cases, the position vs. time graph has a quadratic curve in which we can simply find its acceleration by having initial position and velocity.

In the following, we will try to learn this calculation by a couple of solved examples.

Example (1): The equation of position vs. time for a moving object, in SI units, is as $x=-t^2+6t-9$. Which of the following choices are correct?
(a) The object's acceleration is constant and its magnitude is $1\,{\rm m/s^2}$.
(b) The object's velocity at the initial time $t=0$ is to the negative $x$-axis.
(c) The object's initial position is on the negative side of the $x$-axis.

Solution: we examine each choice separately.

(a) This equation has a quadratic form so its acceleration is constant. Comparing this equation with standard constant acceleration kinematic equation, $x=\frac 12 at^2+v_0t+x_0$, we will find its magnitude as $\frac 12 a=-1 \Rightarrow \ a=-2\,{\rm m/s^2}$ So this choice is incorrect.

(b) Object's velocity at the initial time'' means its initial velocity. Comparing above equations reveals that initial velocity is $v_0=6\,{\rm m/s}$. So, this choice is also incorrect.

(c) By setting $t=0$ in the position-time equation, its initial position is obtained. So, $x_0=-9\,{\rm m}$. Negative indicates that the object is on the negative side of the $x$-axis initially. Thus, this choice is correct.

Example (2): The position vs. time graph of a moving object accelerates uniformly along the $x$ axis is plotted in the figure below. Find its acceleration, initial velocity, and position. Solution: The concavity of the graph tells us about the sign of acceleration. Here, the graph opens upward (concave upward), so its acceleration is positive, $a>0$.

At the instant the motion is started $t=0$, the position of the object is a negative value. It means that initially, the object starts its motion on the negative side of the $x$-axis.

Recall that the slope of the position-time graph represents the object's velocity. A tangent line at time $t=0$ has a negative slope because that makes an obtuse angle with the $+x$-axis. So, the initial velocity is negative. Overall, this object starts its motion at a position behind the origin in the opposite direction of the $x$-axis as shown in the figure below. In the next example, we will find the constant acceleration of a moving object using its position vs. time graph.

Example (3): the position of a moving object (along a straight positive line) as a function of time is given by the curve shown in the figure below. Find the acceleration of the object. Solution: First, collect all information that the plot gives us. At time $t=0$, the object is at position $x=-9\,{\rm m}$ and at time $t=3\,{\rm s}$ its position is zero that is it returns back to its starting position.

The graph opens upward, indicating a positive acceleration.

It is said that the motion has a constant acceleration, so its position versus time must be changed as a quadratic function which is determined by the kinematics equation $x=\frac 12 at^2+v_0t+x_0$.

In the above equation, $x_0$ is the position at time $t=0$ or the initial position. The graph shows us that, for this object, it is $x_0=-9\,{\rm m}$.

$v_0$ is also the initial velocity which is found by computing the slope of the position-time graph at time $t=0$.

As you can see, in this graph, the slope (in green) is parallel to the horizontal, makes an angle of zero, and consequently, its initial velocity is zero, $v_0=0$. If we substitute these two finding into the standard kinematics equation $x=\frac 12 at^2+v_0t+x_0$, we get $x=\frac 12 at^2-9$ The remaining quantity is the acceleration $a$.

There is another point that has not been used yet, $B=(x=0,t=3\,{\rm s})$. Putting this point into the above equation, we get \begin{align*} x&=\frac 12 at^2-9\\\\0&=\frac 12 (a)(3)^2-9\\\\\Rightarrow a&=2\quad {\rm m/s^2}\end{align*} Therefor, the object's acceleration is $2\,{\rm m/s^2}$.

Example (4): A car starts at rest and accelerates at a constant rate in a straight line. Its position-versus-time graph is shown in the figure below. Find the car's acceleration? Solution: This is another example problem that shows you how to find acceleration from a position vs. time graph.

Like previous example, locate two points on the graph with given information. $A(x=-8\,{\rm m},t=0)$ and $B(x=0,t=2\,{\rm s})$. Using these two points and applying the kinematics equation $x=\frac 12 at^2+v_0t+x_0$, one can find the car's acceleration.

Putting point A into the above equation, gives us \begin{align*} x&=\frac 12 at^2+v_0t+x_0\\\\-8&=\frac 12 a(0)^2+v_0(0)+x_0\\\\\Rightarrow \quad x_0&=-8\,{\rm m}\end{align*} In the problem said that the car starts its motion from rest, so its initial velocity is zero, $v_0=0$. Now, putting the second point B into the standard equation, and solving for $a$, get \begin{align*}x&=\frac 12 at^2+v_0t+x_0\\\\0&=\frac 12 a(2)^2+0-8\\\\ 0&=2a-8\\\\\Rightarrow a&=4\quad {\rm m/s^2}\end{align*}

In the next example, we want to identify the type of motion qualitatively, using a position vs. time graph.

Example (5): What type of motion do each of the following position-time curves show?

In the next example, acceleration on a position-time graph gets in the case of having an initial velocity, $v_0\neq0$.

Example (6): The position vs. time graph of a moving object along the positive $x$-axis is as follows. Find the object's acceleration. Solution: As always, to find the constant acceleration of a moving object from its position-versus-time graph, one should locate two points on the graph and substitute them into the standard kinematics equation $x=\frac 12 at^2+v_0t+x_0$.

Here, initial point has coordinate $A(x=18\,{\rm m},t=0)$ from which we can find the initial position, $x_0=18\,{\rm m}$.

The other point is $B(x=0,t=6\,{\rm s})$. Given the initial position, substitute the point B into the standard kinematics equation, we have \begin{gather*} x=\frac 12 at^2+v_0t+x_0\\\\0=\frac 12 a(6)^2+v_0 (6)+18 \\\\ \Rightarrow 18a+6v_0+18=0 \end{gather*} As you can see, we have one equation with two unknowns. We need an additional equation to be able to find the unknowns.

In the previous examples, the position-time graph had a zero slope and thus get a zero initial velocity.

Of the graph, we see that the slope at time $t=0$ is not zero so the object does not start from rest.

But how to find another equation?

To find that equation, we pay attention to an important note below:

The tangent line at point $B$ is horizontal so velocity at that time is zero, according to the equivalence of slope and velocity on a $x-t$ graph.

Substituting this extra information into another kinematics equation, $v=v_0+at$, we can find the initial velocity. \begin{gather*} v=v_0+at\\0=v_0+a(6)\\ \Rightarrow 6a+v_0=0\end{gather*} Solving the above equation for $v_0$, we get $v_0=-6a$.

Now, substitute it into the first equation $18a+6v_0+18=0$ and solve for the unknown acceleration \begin{gather*} 18a+6v_0+18=0\\ 18a+6(-6a)+18=0\\ \Rightarrow a=1\quad {\rm m/s^2}\end{gather*} If we put this value into $v_0=-6a$, the initial velocity is also found.

Consequently, the car starts its motion toward the negative $x$ axis with an initial velocity of 6 m/s and increases its speed at a constant rate of $1\,{\rm m/s^2}$.

Example (7): The position vs. time graph of a moving object along a straight line is a parabola as below. Find the equation of the object's velocity as a function of time? Solution

Physical interpretation of graph: This object starts its motion at some initial velocity (because the slope at time $t=0$ makes an angle with horizontal) and decreases its speed at a constant rate (i.e. $a<0$ as can be seen from the concavity downward of the curve) until reaches point B where its velocity gets zero, changes its direction, and returns to the starting point in the opposite direction.

As you guess, this is exactly a description for motion that appears in a free-fall. In addition, such a graph appears also in the projectile motion problems

As said, the curve of the position-time graph is a parabola that has a quadratic form. So, the object has a constant acceleration.

On the other side, the curve opens downward, so the acceleration is negative $a<0$.

In this curve, the coordinate of three points are given, $A(x=0,t=0)$, $B(x=8\,{\rm m},t=2\,{\rm s})$, and $C(x=0\,{\rm m}, t=4\,{\rm s})$.

The velocity of an object with a constant acceleration changes with time as $v=v_0+at$. Thus, we must find its initial velocity as well as acceleration. To find the unknowns, it is better to apply kinematics equations between any two given points. Between points A and B, displacement is computed as $\Delta x=x_B-x_A=8-0=8\quad{\rm m}$ Using kinematics equation, $v^2-v_0^2=2a\Delta x$, for these two points, we have \begin{gather*} v_B^2-v_A^2=2a\Delta x\\\\0-v_0^2=2(a)(8)\\\\\Rightarrow\quad v_0^2=-16a\end{gather*} where we used this note that velocity at point B is zero because the slope of the tangent line at that point is horizontal.

Now, we choose points A and C, compute their displacement, and apply the kinematics equation $x-x_0=\frac 12 at^2+v_0t$ $\Delta x=x_C-x_A=0-0=0$ As expected since the objects returns to its starting point. \begin{gather*} \Delta x=\frac 12 at^2+v_0t\\\\0=\frac 12 (a)(4)^2+v_0(4)\\\\ \Rightarrow \quad 8a+4v_0=0 \end{gather*} Until now, we have two equations with two unknowns which is completely solvable.

The second equation, $8a+4v_0=0$, gives us $v_0=-2a$. Now, substitute it into the first equation \begin{gather*} v_0^2=-16a\\\\ (-2a)^2=-16a\\\\4a^2=-16a\\\\ \rightarrow\quad 4a(a+4)=0\end{gather*} Solving this equation for $a$, we get two solutions $a=0$, and $a=-4\,{\rm m/s^2}$.

The zero acceleration $a=0$ is not plausible, since it represents a constant velocity motion whose position-time graph is a straight line but in this problem, we have a curve.

Therefore, the acceptable acceleration is $a=-4\,{\rm m/s^2}$. By substituting this into either first ($v_0^2=-16a$) or second equation ($8a+4v_0$), and solving for $v_0$, we will get the initial velocity, \begin{gather*} 8a+4v_0=0 \\ \rightarrow (8)(-4)+4v_0=0\\ \Rightarrow v_0=8\,{\rm m/s}\end{gather*} As expected, since the tangent line at time $t=0$ has a positive slope.

Substituting these known values into the kinematics equation $v=v_0+at$, we will obtain the object's velocity as a function of time. $v=8-4t$ This equation gives the variation of velocity as a function of time. When this equation is plotted, a velocity-time graph is obtained.

## Summary:

In this article, we found out how to compute the object's constant acceleration using a position-time graph.

Given the initial velocity $v_0$, calculate the displacement $\Delta x$ between two known points on the graph. Next, use the kinematics equation $\Delta x=\frac 12 at^2+v_0t$ and solve for the unknown acceleration $a$.

If the object has an initial velocity, then we need at least three points on the graph with known position and time coordinates.

In this case, the kinematics equation $v^2-v_0^2=2a\Delta x$ is also used.

Author: Ali Nemati
Page Published: 8-13-2021