In this explainer, we will learn how to calculate acceleration using an object’s initial velocity, its displacement, and its acceleration time using the formula .
Imagine an object moving in a straight line at a velocity . Let’s say that this object speeds up steadily so that after a time , it moves at a new velocity .
If we choose to let the initial moment in time equal 0, then we can plot points for this object on a velocity–time graph as follows.
We see that over time, the velocity of our object increases. And we are told this increase happens at a constant rate. Therefore, the object accelerates constantly, and since acceleration in general equals a change in velocity divided by a corresponding change in time, we can write
Graphically, the acceleration is equal to the slope of the straight line connecting the points and .
From this graph, we can solve for the total displacement of our object as it speeds up. That displacement equals the area under the curve.
Using the given information about our object, we can calculate its displacement over this time interval. To compute the area under our curve, we can divide that area into a rectangular part and a triangular part as follows.
Letting the variable represent displacement, we write
Since is in the shape of a triangle, it will equal one-half times the triangle’s base multiplied by its height. The triangle’s base is time , and its height is velocity :
, being a rectangle, will equal the base of the rectangle multiplied by its height :
Thus,
Recalling that we can multiply both sides of this equation to yield
Therefore, we can replace in our equation for displacement with as follows:
Rearranging slightly, we obtain the following result.
Formula: Displacement in terms of Initial Velocity, Time Elapsed, and Constant Acceleration
Consider
This result is an equation of motion that describes the displacement of a uniformly accelerating object in terms of its initial velocity , acceleration , and acceleration time .
Note that this is a vector equation. This means the directions of displacement, initial velocity, and acceleration must be taken into account.
This formula applies even when the initial velocity is zero. In that case, the graph of velocity versus time appears as shown below and the relationship for displacement simplifies to
Let’s practice working with the general equation of motion using several examples.
Example 1: Analyzing a Graph of Velocity versus Time
The graph shows the change in velocity of an object with time.
- What is the object’s velocity at ?
- For how long does the object accelerate?
- What is the object’s velocity after it has accelerated?
- What is the object’s acceleration?
- What is the object’s displacement?
Answer
Part 1
Along the horizontal time axis, the line overlaps the vertical (velocity) axis. The velocity at this time is shown to be 20 m/s.
Part 2
Considering how long the object accelerates, we know this equals the length of time over which the object’s velocity changes. The graph shows velocity changing from 0 up to 15 seconds, so our answer is that the object accelerates for 15 seconds.
Part 3
At the end of these 15 seconds, the object’s new velocity is given by the corresponding vertical-axis value. That value is 50 m/s.
Part 4
To solve for the object’s acceleration, we can recall that, in general, where is a change in velocity and is the corresponding change in time.
For the object in question, and
Therefore, the acceleration
Part 5
We can solve for object displacement in two different ways.
First, we can use an algebraic method. Remembering the equation of motion where is the object’s displacement, is its initial velocity, is its acceleration, and is the acceleration time, we can substitute in the known values of , , and as follows:
Another way to solve for this displacement is to determine it graphically. The displacement of our object equals the area under the curve in our graph.
In this figure, we divide the area under the curve into a triangular region and a rectangular region . The total area (and therefore the displacement of our object) is given by the sum of and :
Since the shape corresponding to is a triangle, its area is one-half the triangle’s base multiplied by its height. Looking at the figure above, we see the base is a time of 15 s and the height is a velocity of , or 30 m/s.
Therefore,
Since is a rectangle, its area equals its base (also 15 s) multiplied by its height of 20 m/s:
Therefore,
The displacement of the object over this time interval is 525 m.
We can consider acceleration over a distance in which the initial velocity and final velocity of an object are known and also the time for which the object accelerates.
Let’s now look at an example involving this method.
Example 2: Determining Displacement for an Object Accelerating in the Direction of Its Initial Velocity
An object has an initial velocity of 12 m/s. The object accelerates at 2.5 m/s2 in the same direction of its velocity for a time of 1.5 s. What is the displacement of the object during this time? Answer to one decimal place.
Answer
Since this object accelerates at a steady rate, we can describe its motion using the equations of motion. Specifically, we will use the relationship where is the object’s displacement, is its initial velocity, is its acceleration, and is its acceleration time. Note that displacement, velocity, and acceleration are vectors, indicating that the relative direction of these quantities is important.
Our problem statement tells us that in this instance, all of these vector quantities have the same direction. Modeling the object as a dot, we can choose these vectors with positive quantities to point to the right.
Note that the relative lengths of the three arrows cannot be compared since they indicate different physical quantities.
The diagram above shows that the object’s displacement is in the same direction as its acceleration and initial velocity. Since those values are positive, displacement will be positive as well.
Knowing that , , and ,
Rounding this result to one decimal place, we find that the displacement of the object is 20.8 m.
Example 3: Determining Distance Traveled for a Uniformly Decelerating Car
A car is traveling east and passes the point that is 45 m east of a road junction. As the car passes the point , its speed is 30 m/s and the driver applies the brakes, accelerating west at 2.5 m/s2. What distance east of the road junction is the car 10 s after the brakes are applied?
Answer
A sketch of the car as it passes point could appear as follows.
At the instant shown, the car driver applies the brakes, giving the car a constant deceleration of 2.5 m/s2 to the west, making the car slow down. After 10 s of this deceleration, we want to solve for the car’s total distance east of the intersecting roads.
Because the car’s deceleration is constant over time, we can describe the car’s motion from point using the equation of motion where is the car’s displacement from point , is its initial velocity, is its acceleration, and is the time elapsed.
This equation involves vector quantities, and therefore defining a positive direction and a negative direction in our scenario will be helpful. We choose the eastward direction to be positive, meaning any vector pointing to the west is considered negative. Therefore, the car’s initial velocity is positive, while its acceleration is negative.
Solving for the car’s displacement from point according to the equation of motion above, we have
Here, we must recall that the car started out 45 m east of the intersecting roads. Therefore, the car’s total distance east of the road junction seconds after it passes point is
Let’s now consider a situation in which a moving object accelerates in the direction opposite its initial motion, ending up with a final velocity and displacement also opposite that initial velocity direction.
Example 4: Calculating Object Displacement in the Direction of the Object’s Initial Velocity
An object has an initial velocity of 32 m/s. The object accelerates at 12 m/s2 in the opposite direction to its velocity for a time of 5.5 s. What is the net displacement of the object in the direction of its initial velocity during this time?
Answer
This example involves vector quantities, so it is important to keep careful track of positive and negative signs. We can arbitrarily choose the object’s initial velocity to be toward the right, meaning that its acceleration points to the left, as follows.
In this diagram, we cannot compare the relative lengths of the arrows because they correspond to different physical quantities. We do know however that the object’s net displacement will be positive if it points to the right relative to the start point and negative if it points to the left.
Because our object accelerates at a constant rate, we can use the following equation of motion to describe its movement: where is the object’s displacement, is its initial velocity, is its acceleration, and is the time over which the object accelerates.
We have defined the initial velocity to be positive , so acceleration must therefore be negative ( m/s). Plugging these values into our equation along with time , we get
The object’s net displacement in the direction of its initial velocity is m, telling us the net displacement arrow in the figure above should point to the left.
Lastly, we consider an example where we solve for object velocity.
Example 5: Calculating Initial Velocity in the Direction of Acceleration
A car that was initially moving at a steady speed has a net displacement of 45 m after accelerating in a straight line at 1.5 m/s2 for 15 seconds. What was the car’s initial velocity in the direction of its acceleration?
Answer
Because displacement, acceleration, and velocity are all vector quantities, we must be careful to understand the relative directions in which they act.
Note that the given displacement (45 m) and acceleration (1.5 m/s2) are both positive, telling us that these vectors point in the same direction. We want to solve for the car’s initial velocity relative to the sign convention that motion in the direction of the car’s acceleration is positive.
Since the car accelerates at a constant rate over the 15 s time interval, we can use the following equation of motion to describe its movement: where is the car’s displacement, is its initial velocity, is its acceleration, and is the acceleration time.
We can rearrange this equation to solve for initial velocity . Subtracting from both sides gives
Dividing both sides of the equation by gives so that
Substituting in the known values for (45 m), (1.5 m/s2), and (15 s), we have
The fact that is negative means it acts in the direction opposite the car’s acceleration. Relative to that positive direction then, the car’s initial velocity is m/s.
Key Points
- The equation of motion can be derived algebraically or from a graph of object motion under constant acceleration.
- The quantities (displacement), (initial velocity), and (acceleration) in this equation are all vectors and therefore can be positive or negative.
- This equation can be arranged algebraically so that any of the variables (, , , or ) is the subject.
