Kinematics II: _{Velocity and acceleration in one dimension}

Vector quantities and reference frames

To communicate clearly about the motion of an object like a rocket, physicists and engineers first describe a reference frame for the motion. A “reference frame” defines an origin and a positive direction for the position of an object. For the motion of a New Shepard rocket, the reference frame defines the origin as Earth’s surface where the launch takes place, and the positive direction is perpendicular to Earth’s surface or up (see Figure 2).

In this module, we will focus on one-dimensional motion and ignore the rocket's horizontal motion, so this is a one-dimensional reference frame.

The rocket's base is located at the origin, so it is described as having a position of 0, as shown in Figure 2. The bottom of the capsule that holds the passengers is 15 meters off the ground, so it is described as having a position of +15m, which we read as “positive fifteen meters.” The “+” is an important part of the position, because position is a vector quantity. “Vector quantities” have both a magnitude and a direction. In one dimension, a “+” or a “-” describes the direction of the vector as either “up” or “down”. In this reference frame, negative positions would be located under the ground.

“Velocity” is also a vector quantity closely related to the scalar quantity speed. However, velocity describes the rate of change of the rocket's position and includes the direction. In a one-dimensional reference frame, the direction will be up or down, as with position. A scientist may choose to state the direction or to use a “+” or “-” sign. For example, the landing velocity of the New Shepard rocket is 2.7m/s downward or -2.7m/s.

“Acceleration” is the third vector quantity physicists use to describe motion. It describes how the velocity is changing. An upward acceleration describes a velocity that is becoming more positive, like when the rocket is taking off and moving upward faster and faster. This module will discuss acceleration in detail.

Describing motion in one dimension

Figure 3 shows the flight plans for the New Shepard rocket and capsule in three main stages.

1. In the first stage, the rocket and capsule accelerate upward together.
2. In the second stage, the rocket and capsule separate while traveling upward, reaching the highest point in their trajectories before falling back toward Earth.
3. In the final stage, the rocket and capsule land gently.

The rocket uses its engine to slow down, while the capsule uses a parachute. Since the rocket reaches a maximum height of about 100km above the launch site and lands just 3.2km away, the motion is very nearly one-dimensional.

The one-dimensional vertical motion of the New Shepard rocket changes slightly each time it is launched, depending on its payload and mission. It is a complex motion that physicists describe by breaking the motion into different time intervals. Figure 4 is the position vs. time graph representing the full motion of the New Shepard rocket in five pieces from launch to landing for the April 14, 2025, mission.

Physicists add detail to the graph-based description using scalar and vector quantities that describe an object's motion. For example, when the rocket reaches a position of +45,300m and is traveling at +1000 m/s, the rocket engine turns off.

Describing change: instantaneous velocity

One goal of the Blue Origin design team is to launch the New Shepard space capsule above the Kármán line, a generally agreed-upon “edge of outer space.” At 100km above Earth’s surface, the Kármán line is where the atmosphere becomes too thin to create enough lift for conventional aircraft, like commercial airplanes, to operate (see Figure 3). The Kármán line is named after its founder, Theodore von Kármán, a Hungarian-born aerospace engineer who emigrated to America in the 1930s and co-founded several rocket-focused businesses and research facilities, including Aerojet and the Jet Propulsion Laboratory (JPL). Today, we seek to send spacecraft and satellites beyond this limit.

To reach the Kármán line (in other words, to change its position), the New Shepard rocket must change its velocity. The rocket remains at rest for seven seconds after the engine ignites, then moves slowly upwards. After 10 seconds (10s), the rocket has just cleared the launch tower. At 20s, the rocket has a velocity of +60m/s and is at a height of +0.34km, and at 60s has a velocity of +250m/s and a height of +7.0km. When the engine is turned off at 140s, the rocket is moving at +1000m/s and is barely visible from the ground at a height of +45.0km. The position vs. time graph (Figure 5, panel A) shows the same data as the first 50s shown in Figure 4. During this time, the rocket travels approximately +5km vertically.

In addition to position vs. time graphs, motion diagrams (Figure 5, panel B) can organize and represent motion graphically. Motion diagrams will include the axis of the reference frame with the object’s position represented as points along the axis. These points are labeled when multiple points are used to show different moments in time. The vector quantities are added to the motion diagram as arrows originating from the points in the case of velocity. In the motion diagram in Figure 5, Panel B, the changing velocity is represented as velocity vectors with changing length. The velocity is greater during each 10s time interval, and the velocity vector is longer.

The same change can be seen in the position vs. time graph in Figure 5, Panel A. Every 5s, a green point shows the rocket's position. The green tangent line going through each point shows the slope of the best fit curve, also called the “instantaneous slope.” The tangent line's changing slope matches the changing velocity; the slope becomes steeper as the velocity increases.

Velocity is defined as the rate of change of position with respect to time. As an equation, it’s represented as \( v = \frac{\Delta x}{\Delta t} \). In this equation, the x is the “displacement” of the object for the time interval t. Displacement is the vector quantity representing the change in an object’s position. Therefore, finding a value for an object’s velocity requires knowing where the object is at two different moments in time, just like finding the slope of a line on a graph requires two different points on the line. This works well for motion with a constant velocity, as illustrated by the data in Figure 6 when the rocket engine is cut off.

In Figure 6, during each 0.2s time interval, the rocket has a displacement of +200m. When calculating the rocket's velocity at 140s, we can choose a time interval of 0.2s and a displacement of +200m and calculate the velocity as:

If we choose a time interval of 0.4s and a displacement of +400m. The result is the same: the velocity is +1000 m/s.

In the graph of Figure 6, the constant velocity is seen as a constant slope for the graph. You can look at ou module on Linear Equations for a more detailed treatment of linear relationships.

When the velocity is not constant, as in Figure 5, the value of the velocity depends on the choice of time interval, which was a problem for ancient philosophers like Aristotle (384 BCE - 322 BCE) and early scientists like Galileo Galilei (1564-1642). When finding the rocket's velocity at 35s, if the time interval is chosen as 1.0s, the velocity value is +140m/s. If a shorter time interval of 0.2s is used, the velocity value is slightly smaller at +138m/s. Shorter and shorter time intervals give the velocity values that converge at +137m/s.

Isaac Newton, a British mathematician and scientist, developed the mathematical field of calculus, focusing on the mathematics of continuous change. In Newton’s 1687 work Principia Mathematica, he provided the mathematical basis for the concept of “instantaneous velocity” by defining the slopes of the tangent lines to curves as the rate of change of a mathematical function. Newton argued that because the slope of the tangent line at each moment in time is a well-defined property of the position vs. time graph (as shown in Figure 5A), the instantaneous velocity is also a well-defined property of motion at each moment in time. Today, when physicists use the term “velocity,” they typically mean the instantaneous velocity of an object, which changes throughout its motion.

Another way of representing the rocket’s motion is to graph the instantaneous velocity. Figure 7 shows the same motion over time as Figure 4, but graphs the velocity of the New Shepard rocket on the y-axis instead of its position. The same periods of launch, free-fall, braking, and landing are shown.

Velocity vs. time graphs can be less intuitive to interpret than position vs. time graphs. While both graphs appear the same initially, with 7 seconds of data on the horizontal axis, these data have different meanings. Zero on the position vs. time graph indicates that the object is at the origin; for the rocket, this happens at the beginning and the end of the motion. Zero on the velocity vs. time graph indicates that the object is not moving. For the rocket, this also happens at the motion's start and end. Additionally, there is a moment near 250s when the rocket has zero velocity, which is when it reaches the highest point, as seen in Figure 4.

The position vs. time graph for the rocket is positive for all moments during the motion (Figure 4). This is because the origin has been defined at ground level, and the rocket always stays above this point. However, the velocity vs. time graph includes positive values when the rocket moves upward and negative values when the rocket moves downward (Figure 7). Near 250s, when the velocity changes from positive to negative, the rocket stops moving away from the surface of Earth and starts moving towards it.

Note in Figure 7 that, between 130s and 140s, the rocket's velocity remains constant at 1000 m/s. This is when the rocket engine cuts off and no longer adds thrust. In the position vs. time graph (Figure 4), a constant slope indicates a constant velocity; the value of the slope is equal to the velocity. A constant velocity appears as a horizontal line on the velocity vs. time graph. Most of the velocity vs. time graph shows neither zero nor constant velocity. Instead, the velocity is constantly changing, which physicists call acceleration.

Describing and calculating acceleration

The velocity vs. time graph in Figure 7 shows that the rocket’s velocity constantly changes. “Acceleration” describes the rate at which an object’s velocity changes over time. Acceleration, represented by the variable \(a\), is defined mathematically as

The Δv stands for the change in the object's velocity. It can also be written as the difference between the velocity at the end of the time interval (final velocity, v) and velocity at the start of the time interval (initial velocity, v₀). The units of acceleration will be the units of the velocity over the unit of time—in many physics applications, this will be meters per second per second, or m/s².

In everyday language, acceleration refers to speeding up, and deceleration refers to slowing down. Physicists use the word acceleration to describe both speeding up and slowing down and use positive and negative signs to show the direction in which the velocity is changing. The motion diagrams in Figure 8 illustrate several relationships between velocity and acceleration.

• Panel A shows an object with a constant positive velocity (the green arrows are all the same length in the positive direction) and thus zero acceleration. The displacement between successive positions is also constant. This motion diagram matches the engine cutoff for the New Shepard rocket.
• Panel B shows an object with a positive initial velocity and a positive acceleration. As the acceleration points in the same direction as the velocity, it reinforces the velocity, increasing its value. In each second of motion, the velocity is becoming more positive by the same amount, and the object is speeding up. The displacement between successive positions is increasing along with the velocity. This motion diagram matches the rocket powered launch for the New Shepard rocket.
• Panel C shows an object with a positive initial velocity and a negative acceleration. As the acceleration points in the opposite direction of the velocity, it opposes the velocity, decreasing its value. During each second of motion, the velocity is becoming smaller by the same amount. In this case, the object is slowing down, and the displacement between successive positions is decreasing. This motion diagram matches the rising free-fall motion for the New Shepard rocket.
• Panel D shows an object with a negative initial velocity and a negative acceleration. The velocity and acceleration vectors point in the same direction, so the acceleration reinforces the velocity. During each second of motion, the velocity is becoming more negative by the same amount. In this case, the object is speeding up, but the acceleration is negative. This motion diagram matches the falling free-fall motion for the New Shepard rocket.
• Panel E shows an object with a negative initial velocity and a positive acceleration. Most people take extra care with this example. Similar to Panel C, the acceleration opposes the velocity. Since the initial velocity is in the negative direction, the positive acceleration reduces the magnitude of the negative velocity, so the object is slowing down. The velocity is becoming less negative (more positive) so that after 3s, the object has come to rest. This example highlights how the sign (“+” or “-”) corresponds to a direction for the acceleration and not to whether the object is speeding up or slowing down. This motion diagram matches the atmospheric braking and rocket powered landing motion for the New Shepard rocket.

Figure 9 shows two graphical representations of acceleration during the New Shepard rocket's liftoff. Since acceleration is defined as the rate of change of velocity, it is easiest to see on a velocity vs. time graph (see Figure 9, Panel A). The acceleration is simply the slope of the velocity vs. time graph. During the liftoff, the acceleration was +5.4 m/s².

Accelerated motion appears as a curve on a position vs. time graph (Figure 9, Panel B and Figure 5A). Because the velocity is constantly changing, the slope of the position vs. time graph is also constantly changing. The position is no longer a linear function of time as it is when the velocity is constant. Instead, when the acceleration is constant, the position is a function of the square of time \((t^2)\) . This is described as a quadratic dependence, and the position vs. time graph will have a parabolic shape. The quadratic dependence appears as a term with \(t^2\) in the equation for the object's position at time t, as seen in Table 1. One way of understanding the \(t^2\) term is that the position (\(x\)) is proportional to time and velocity \(x \propto v \cdot t\). Since the velocity is proportional to time when the acceleration is constant \(x \propto a \cdot t\), the distance is proportional to \(t^2\)

Using mathematics, physicists can describe how initial position, initial velocity, and a constant acceleration combine to determine an object’s value of velocity and position. These equations are shown below in Table 1. Constant velocity motion is a special case when the acceleration is zero, and the equations for position and time are simpler.

With the addition of quantitative information, these “kinematic equations” relating position, velocity, acceleration, and time allow physicists to calculate additional information. Two examples illustrate this below.

Example 1: Engine cutoff The rocket has zero acceleration during the engine cutoff, so the equations for constant velocity motion apply. At 140s after engine ignition, we know that the velocity \((v)\) is +1000m/s and the initial position \((x^0)\) is +45300m. The equation \(x = x_0 + v \cdot t\) that applies for constant velocity motion can be used to compute where the rocket will be at the end of a time interval. If we want to know where the rocket is at 140.5s, we choose the time interval t = 0.5s. Using this data, the final position can be calculated.

Example 2: Acceleration at liftoff During the liftoff, the rocket accelerates 5.4m/s², so the equations for constant acceleration apply. At 9s after engine ignition, the rocket begins to accelerate from rest at ground level, so both the initial velocity \((v^0)\) and the initial position are zero at 9s. The equation \(x = x_0 + v_0 t + \tfrac{1}{2} a t^2\) can compute where the rocket will be at the end of a time interval. If we want to know where the rocket is at 50s, we choose the time interval t =50s-9s = 41s . The position at 50s is then calculated.

Similarly, we can predict the velocity at the end of the time interval using the equation \(v = v_0 + a \cdot t\).

The relationship between acceleration and free-fall motion

Italian astronomer and physicist Galileo Galilei experimented extensively with one common and important example of accelerated motion, the motion of an object in free-fall. “Free-fall” is the motion of an object when gravity is the only force acting and air resistance is small enough to be ignored. In free-fall motion, gravity produces a constant acceleration of -9.8m/s² for all objects close to Earth's surface. For the New Shepard rocket, the longest part of the motion from approximately 140s to 370s after ignition can be described as free-fall motion. In fact, for the launch on April 14, 2025, the passengers were excited for this part of the trip because during the several minutes of free-fall, they would experience weightlessness.

Panel A of Figure 10 shows the position vs. time graph for the New Shepard rocket’s period of free-fall motion. Similar to the takeoff period shown in Figure 9, the position vs. time graph is curved, showing that the instantaneous velocity is changing. However, in this case, the position vs. time graph has a positive slope at 150s, showing that it begins with an upward velocity. At 246s, the position reaches a maximum, and the tangent line to the graph is horizontal, indicating zero velocity. From this time onward, the slope of the graph becomes increasingly negative, showing that the velocity is downward and the speed is increasing.

In Panel B of Figure 10, we also see an initially positive velocity of ~900 m/s at 150s, zero velocity at 246s, and an increasingly negative velocity at longer times. Additionally, the data fit well with a linear function with a slope of -9.5m/s2. This indicates that the motion has a constant acceleration downward. Even when the rocket’s velocity reaches zero at the top of the motion, the velocity changes at the same rate. Even at rest, the rocket is accelerating downward.

That acceleration (-9.5m/s²) might sound almost familiar: It is slightly less than the value Galileo learned is the acceleration due to gravity's force at Earth's surface (9.8m/s²). In fact, the force of gravity and the acceleration due to gravity change as objects move away from the planet's surface, as Newton’s Law of Universal Gravitation predicted, which you can read about in our module on Gravity. This decrease of about 3% at an altitude of 100km is expected.

It may seem contradictory that astronauts who are experiencing weightlessness during their flight are also described as accelerating under the force of gravity. Weightlessness is often confused with the absence of gravity, but this is a misconception based on our perception. We generally experience the force of gravity when another force balances it—the force on our feet when we are standing or on our backs when we lie in bed. When there is no opposing force—for example, if we were in free-fall like the astronauts in the New Shepard rocket—we no longer perceive the force of gravity. This effect is particularly strong for astronauts in a capsule because all the objects around them are also in free-fall, as shown in the photo in Figure 11.

Planning flight parameters for landing the New Shepard rocket

The most challenging aspect of designing the flight parameters for the New Shepard rocket is the landing and the final minute of flight, shown in Figure 12. As the rocket approaches the ground, it is traveling at a maximum negative velocity near -1200m/s, more than four times the speed of sound at Earth’s surface. Near 40km above the surface and 360s after launch, the rocket begins to encounter enough air at such high speeds that air resistance is no longer negligible. At some point, the rocket reaches a terminal velocity of -160m/s. This occurs when the force of air resistance balances the force of gravity, and the rocket's acceleration goes to zero. The velocity of the rocket then remains constant from approximately 420s to 430s, which is visible in panel A of Figure 12.

The final landing of the rocket cannot be handled through air resistance alone—it would crash, like most previous rocket engines. Instead, a gentle landing is accomplished by re-igniting the engine, creating positive acceleration that reduces the velocity of the rocket to zero precisely when the rocket reaches ground level.

Flight engineers have two parameters to consider when planning for a successful landing. First, they can program when to reignite the engine. Choosing this moment will determine the altitude and velocity the rocket will have as it begins the landing process. Second, they can adjust the rocket's acceleration by changing the engine's burn rate. The equations describing the final motion are given by:

In reality, the flight planning for landing a rocket is more complex. In this simulation, horizontal motion is ignored, which is very important for the New Shepard rocket when it lands. Additionally, the rocket's orientation must be maintained straight up and down. Each of these requirements poses additional constraints on the kinematics of the rocket’s motion and requires additional design elements in the New Shepard rocket.