Lesson Explainer: Acceleration over Distance and Time Physics • 9th Grade

In this explainer, we will learn how to calculate acceleration using an objectโ€™s initial velocity, its displacement, and its acceleration time using the formula ๐‘ =๐‘ข๐‘ก+12๐‘Ž๐‘ก๏Šจ.

Imagine an object moving in a straight line at a velocity ๐‘ข. Letโ€™s say that this object speeds up steadily so that after a time ๐‘ก, it moves at a new velocity ๐‘ฃ.

If we choose to let the initial moment in time equal 0, then we can plot points for this object on a velocityโ€“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 ๐‘Ž=๐‘ฃโˆ’๐‘ข๐‘ก.

Graphically, the acceleration is equal to the slope of the straight line connecting the points (0,๐‘ข) and (๐‘ก,๐‘ฃ).

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 ๐‘  represent displacement, we write ๐‘ =+.areaarea๏Šง๏Šจ

Since area๏Šง is in the shape of a triangle, it will equal one-half times the triangleโ€™s base multiplied by its height. The triangleโ€™s base is time ๐‘ก, and its height is velocity ๐‘ฃโˆ’๐‘ข: area๏Šง=๏€ผ12๏ˆร—๐‘กร—(๐‘ฃโˆ’๐‘ข).

Area๏Šจ, being a rectangle, will equal the base of the rectangle (๐‘ก) multiplied by its height (๐‘ข): area๏Šจ=๐‘กร—๐‘ข.

Thus, ๐‘ =๏€ผ12๏ˆร—๐‘กร—(๐‘ฃโˆ’๐‘ข)+๐‘กร—๐‘ข.

Recalling that ๐‘Ž=๐‘ฃโˆ’๐‘ข๐‘ก, we can multiply both sides of this equation to yield ๐‘Žร—๐‘ก=๐‘ฃโˆ’๐‘ข.

Therefore, we can replace ๐‘ฃโˆ’๐‘ข in our equation for displacement with ๐‘Žร—๐‘ก as follows: ๐‘ =๏€ผ12๏ˆร—๐‘กร—(๐‘Žร—๐‘ก)+๐‘กร—๐‘ข.

Rearranging slightly, we obtain the following result.

Formula: Displacement in terms of Initial Velocity, Time Elapsed, and Constant Acceleration

Consider ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก.๏Šจ

This result is an equation of motion that describes the displacement of a uniformly accelerating object in terms of its initial velocity ๐‘ข, acceleration ๐‘Ž, and acceleration time ๐‘ก.

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 ๐‘ข is zero. In that case, the graph of velocity versus time appears as shown below and the relationship for displacement ๐‘  simplifies to ๐‘ =๏€ผ12๏ˆร—๐‘Žร—๐‘ก(๐‘ข=0).๏Šจ

Letโ€™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.

  1. What is the objectโ€™s velocity at ๐‘ก=0?
  2. For how long does the object accelerate?
  3. What is the objectโ€™s velocity after it has accelerated?
  4. What is the objectโ€™s acceleration?
  5. What is the objectโ€™s displacement?

Answer

Part 1

Along the horizontal time axis, the line ๐‘ก=0 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 objectโ€™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 objectโ€™s new velocity is given by the corresponding vertical-axis value. That value is 50 m/s.

Part 4

To solve for the objectโ€™s acceleration, we can recall that, in general, ๐‘Ž=ฮ”๐‘ฃฮ”๐‘ก, where ฮ”๐‘ฃ is a change in velocity and ฮ”๐‘ก is the corresponding change in time.

For the object in question, ฮ”๐‘ฃ=50/โˆ’20/=30/msmsms and ฮ”๐‘ก=15.s

Therefore, the acceleration ๐‘Ž=30/15=2/.mssms๏Šจ

Part 5

We can solve for object displacement in two different ways.

First, we can use an algebraic method. Remembering the equation of motion ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก,๏Šจ where ๐‘  is the objectโ€™s displacement, ๐‘ข is its initial velocity, ๐‘Ž is its acceleration, and ๐‘ก is the acceleration time, we can substitute in the known values of ๐‘ข, ๐‘ก, and ๐‘Ž as follows: ๐‘ =(20/)ร—(15)+๏€ผ12๏ˆร—๏€น2/๏…ร—(15)=300+225=525.mssmssmmm๏Šจ๏Šจ

Another way to solve for this displacement is to determine it graphically. The displacement ๐‘  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 (๐ด)๏Šง and a rectangular region (๐ด)๏Šจ. The total area (and therefore the displacement of our object) is given by the sum of ๐ด๏Šง and ๐ด๏Šจ: ๐‘ =๐ด+๐ด.๏Šง๏Šจ

Since the shape corresponding to ๐ด๏Šง is a triangle, its area is one-half the triangleโ€™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 50/โˆ’20/msms, or 30 m/s.

Therefore, ๐ด=๏€ผ12๏ˆร—(15)ร—(30/)=225.๏Šงsmsm

Since ๐ด๏Šจ is a rectangle, its area equals its base (also 15 s) multiplied by its height of 20 m/s: ๐ด=(15)ร—(20/)=300.๏Šจsmsm

Therefore, ๐‘ =225+300=525.mmm

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.

Letโ€™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 ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก,๏Šจ where ๐‘  is the objectโ€™s displacement, ๐‘ข is its initial velocity, ๐‘Ž is its acceleration, and ๐‘ก 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 objectโ€™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 ๐‘ข=12/ms, ๐‘Ž=2.5/ms๏Šจ, and ๐‘ก=1.5s, ๐‘ =(12/)ร—(1.5)+๏€ผ12๏ˆร—๏€น2.5/๏…ร—(1.5)=18+2.8125=20.8125.mssmssmmm๏Šจ๏Šจ

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 ๐‘ƒ that is 45 m east of a road junction. As the car passes the point ๐‘ƒ, 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 ๐‘ƒ 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 carโ€™s total distance east of the intersecting roads.

Because the carโ€™s deceleration is constant over time, we can describe the carโ€™s motion from point ๐‘ƒ using the equation of motion ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก,๏Šจ where ๐‘  is the carโ€™s displacement from point ๐‘ƒ, ๐‘ข is its initial velocity, ๐‘Ž is its acceleration, and ๐‘ก 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 carโ€™s initial velocity ๐‘ข is positive, while its acceleration ๐‘Ž is negative.

Solving for the carโ€™s displacement from point ๐‘ƒ according to the equation of motion above, we have ๐‘ =(30/)ร—(10)+๏€ผ12๏ˆร—๏€นโˆ’2.5/๏…ร—(10)=300+(โˆ’125)=175.mssmssmmm๏Šจ๏Šจ

Here, we must recall that the car started out 45 m east of the intersecting roads. Therefore, the carโ€™s total distance east of the road junction ten seconds after it passes point ๐‘ƒ is 175+45=220.mmm

Letโ€™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 Objectโ€™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 objectโ€™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 objectโ€™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: ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก,๏Šจ where ๐‘  is the objectโ€™s displacement, ๐‘ข is its initial velocity, ๐‘Ž is its acceleration, and ๐‘ก is the time over which the object accelerates.

We have defined the initial velocity ๐‘ข to be positive (+32/)ms, so acceleration ๐‘Ž must therefore be negative (โˆ’12 m/s). Plugging these values into our equation along with time ๐‘ก=5.5s, we get ๐‘ =(32/)ร—(5.5)+๏€ผ12๏ˆร—๏€นโˆ’12/๏…ร—(5.5)=176+(โˆ’181.5)=โˆ’5.5.mssmssmmm๏Šจ๏Šจ

The objectโ€™s net displacement in the direction of its initial velocity is โˆ’5.5 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 carโ€™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 carโ€™s initial velocity relative to the sign convention that motion in the direction of the carโ€™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: ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก,๏Šจ where ๐‘  is the carโ€™s displacement, ๐‘ข is its initial velocity, ๐‘Ž is its acceleration, and ๐‘ก is the acceleration time.

We can rearrange this equation to solve for initial velocity ๐‘ข. Subtracting ๏€ผ12๏ˆร—๐‘Žร—๐‘ก๏Šจ from both sides gives ๐‘ โˆ’๏€ผ12๏ˆร—๐‘Žร—๐‘ก=๐‘ขร—๐‘ก.๏Šจ

Dividing both sides of the equation by ๐‘ก gives ๐‘ โˆ’๏€ป๏‡ร—๐‘Žร—๐‘ก๐‘ก=๐‘ข๏Šง๏Šจ๏Šจ so that ๐‘ข=๐‘ โˆ’๏€ป๏‡ร—๐‘Žร—๐‘ก๐‘ก.๏Šง๏Šจ๏Šจ

Substituting in the known values for ๐‘  (45 m), ๐‘Ž (1.5 m/s2), and ๐‘ก (15 s), we have ๐‘ข=(45)โˆ’๏€ป๏‡ร—๏€น1.5/๏…ร—(15)15=โˆ’8.25/.mmsssms๏Šง๏Šจ๏Šจ๏Šจ

The fact that ๐‘ข is negative means it acts in the direction opposite the carโ€™s acceleration. Relative to that positive direction then, the carโ€™s initial velocity is โˆ’8.25 m/s.

Key Points

  • The equation of motion ๐‘ =๐‘ขร—๐‘ก+๏€ผ12๏ˆร—๐‘Žร—๐‘ก๏Šจ can be derived algebraically or from a graph of object motion under constant acceleration.
  • The quantities ๐‘  (displacement), ๐‘ข (initial velocity), and ๐‘Ž (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 (๐‘ , ๐‘ข, ๐‘Ž, or ๐‘ก) is the subject.

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