2.4.3 The Constant-Acceleration Model

Few objects with changing velocity have a perfectly constant acceleration, but it is often reasonable to model their acceleration as being constant. We do so by utilizing the constant-acceleration model. Once again, a model is a set of words, pictures, graphs, and equations that allows us to explain and predict an object’s motion.

MODEL 2.2 Constant acceleration

For motion with constant acceleration.

Model the object as a particle moving in a straight line with constant acceleration.

Full Alternative Text
Mathematically:
- $v_{f s} = v_{i s} + a_{s} Δ t$
- $s_{f} = s_{i} + v_{i s} Δ t + \frac{1}{2} a_{s} {(Δ t)}^{2}$
- $v_{f s}^{2} = v_{i s}^{2} + 2 a_{s} Δ s$
Limitations: Model fails if the particle’s acceleration changes.

Workbook Exercise 16

In this text, we’ll usually model runners, cars, planes, and rockets as having constant acceleration. Their actual acceleration is often more complicated (for example, a car’s acceleration gradually decreases rather than remaining constant until full speed is reached), but the mathematical complexity of dealing with realistic accelerations would detract from the physics we’re trying to learn.

The constant-acceleration model is the basis for a problem-solving strategy.

PROBLEM-SOLVING STRATEGY 2.1
Kinematics with constant acceleration

MODEL Model the object as having constant acceleration.

VISUALIZE Use different representations of the information in the problem.

Draw a pictorial representation. This helps you assess the information you are given and starts the process of translating the problem into symbols.
Use a graphical representation if it is appropriate for the problem.
Go back and forth between these two representations as needed.

SOLVE The mathematical representation is based on the three kinematic equations:

\begin{array}{l} v_{f s} = v_{i s} + a_{s} Δ t \\ s_{f} = s_{i} + v_{i s} Δ t + \frac{1}{2} a_{s} {(Δ t)}^{2} \\ v_{f s}^{2} = v_{i s}^{2} + 2 a_{s} Δ s \end{array}

Use $x$ or $y$ , as appropriate to the problem, rather than the generic $s$ .
Replace i and f with numerical subscripts defined in the pictorial representation.

REVIEW Check that your result has the correct units and significant figures, is reasonable, and answers the question.

NOTE You are strongly encouraged to solve problems on the Dynamics Worksheets found at the back of the Student Workbook. These worksheets will help you use the Problem-Solving Strategy and develop good problem-solving skills.

EXAMPLE 2.10 The motion of a rocket sled

A rocket sled’s engines fire for 5.0 s, boosting the sled to a speed of 250 m/s. The sled then deploys a braking parachute, slowing by 3.0 m/s per second until it stops. What is the total distance traveled?

MODEL We’re not given the sled’s initial acceleration, while the rockets are firing, but rocket sleds are aerodynamically shaped to minimize air resistance and so it seems reasonable to model the sled as a particle undergoing constant acceleration.

VISUALIZE FIGURE 2.23 shows the pictorial representation. We’ve made the reasonable assumptions that the sled starts from rest and that the braking parachute is deployed just as the rocket burn ends. There are three points of interest in this problem: the start, the change from propulsion to braking, and the stop. Each of these points has been assigned a position, velocity, and time. Notice that we’ve replaced the generic subscripts i and f of the kinematic equations with the numerical subscripts 0, 1, and 2. Accelerations are associated not with specific points in the motion but with the intervals between the points, so acceleration $a_{0 x}$ is the acceleration between points 0 and 1 while acceleration $a_{1 x}$ is the acceleration between points 1 and 2. The acceleration vector $\vec{a}_{1}$ points to the left, so $a_{1 x}$ is negative. The sled stops at the end point, so $v_{2 x} = 0 m/s$ .

A pictorial representation of the movement of a rocket sled after deploying a braking parachute and its motion diagram.

FIGURE 2.23 Full Alternative Text

SOLVE We know how long the rocket burn lasts and the velocity at the end of the burn. Because we’re modeling the sled as having uniform acceleration, we can use the first kinematic equation of Problem-Solving Strategy 2.1 to write

v_{1 x} = v_{0 x} + a_{0 x} (t_{1} - t_{0}) = a_{0 x} t_{1}

We started with the complete equation, then simplified by noting which terms were zero. Solving for the boost-phase acceleration, we have

a_{0 x} = \frac{v_{1 x}}{t_{1}} = \frac{250 m/s}{5.0 s} = 50 {m/s}^{2}

Notice that we worked algebraically until the last step—a hallmark of good problem-solving technique that minimizes the chances of calculation errors. Also, in accord with the significant figure rules of Chapter 1, ${50 m/s}^{2}$ is considered to have two significant figures.

Now we have enough information to find out how far the sled travels while the rockets are firing. The second kinematic equation of Problem-Solving Strategy 2.1 is

\begin{matrix} x_{1} & = & x_{0} + v_{0 x} (t_{1} - t_{0}) + \frac{1}{2} a_{0 x} {(t_{1} - t_{0})}^{2} = \frac{1}{2} a_{0 x} t_{1}^{2} \\ = & \frac{1}{2} (50 {m/s}^{2}) {(5.0 s)}^{2} = 625 m \end{matrix}

The braking phase is a little different because we don’t know how long it lasts. But we do know both the initial and final velocities, so we can use the third kinematic equation of Problem-Solving Strategy 2.1:

v_{2 x}^{2} = v_{1 x}^{2} + 2 a_{1 x} Δ x = v_{1 x}^{2} + 2 a_{1 x} (x_{2} - x_{1})

Notice that $Δ x$ is not $x_{2};$ it’s the displacement $(x_{2} - x_{1})$ during the braking phase. We can now solve for $x_{2} :$

\begin{array}{l} x_{2} & = x_{1} + \frac{v_{2 x}^{2} - v_{1 x}^{2}}{2 a_{1 x}} \\ = 625 m + \frac{0 - {(250 m/s)}^{2}}{2 (- 3.0 {m/s}^{2})} = 11, 000 m \end{array}

We kept three significant figures for $x_{1}$ at an intermediate stage of the calculation but rounded to two significant figures at the end.

REVIEW The total distance is $11 km \approx 7 mi$ . That’s large but believable. Using the approximate conversion factor $1 m/s \approx 2 mph$ from Table 1.5, we see that the top speed is $\approx 500 mph$ . It will take a long distance for the sled to gradually stop from such a high speed.

EXAMPLE 2.11 A two-car race

Fred is driving his Volkswagen Beetle at a steady $20 m / s$ when he passes Betty sitting at rest in her Porsche. Betty instantly begins accelerating at $5.0 {m/s}^{2}$ . How far does Betty have to drive to overtake Fred?

MODEL Model the VW as a particle in uniform motion and the Porsche as a particle with constant acceleration.

VISUALIZE FIGURE 2.24 is the pictorial representation. Fred’s motion diagram is one of uniform motion, while Betty’s shows uniform acceleration. Fred is ahead in frames 1, 2, and 3, but Betty catches up with him in frame 4. The coordinate system shows the cars with the same position at the start and at the end—but with the important difference that Betty’s Porsche has an acceleration while Fred’s VW does not.

The figure depicts the movement of two cars traveling in the same direction, their motion diagrams, and their position-versus-time graphs.

FIGURE 2.24 Full Alternative Text

SOLVE This problem is similar to Example 2.2, in which Bob and Susan met for lunch. As we did there, we want to find Betty’s position ${(x_{1})}_{B}$ at the instant $t_{1}$ when ${(x_{1})}_{B} = {(x_{1})}_{F}$ . We know, from the models of uniform motion and uniform acceleration, that Fred’s position graph is a straight line but Betty’s is a parabola. The position graphs in FIGURE 2.24 show that we’re solving for the intersection point of the line and the parabola.

Fred’s and Betty’s positions at $t_{1}$ are

\begin{array}{l} {(x_{1})}_{F} = {(x_{0})}_{F} + {(v_{0 x})}_{F} (t_{1} - t_{0}) = {(v_{0 x})}_{F} t_{1} \\ {(x_{1})}_{B} = {(x_{0})}_{B} + {(v_{0 x})}_{B} (t_{1} - t_{0}) + \frac{1}{2} {(a_{0 x})}_{B} {(t_{1} - t_{0})}^{2} = \frac{1}{2} {(a_{0 x})}_{B} t_{1}^{2} \end{array}

By equating these,

{(v_{0 x})}_{F} t_{1} = \frac{1}{2} {(a_{0 x})}_{B} t_{1}^{2}

we can solve for the time when Betty passes Fred:

\begin{array}{l} t_{1} [\frac{1}{2} {(a_{0 x})}_{B} t_{1} - {(_{​ ​} v_{0 x})}_{F}] = 0 \\ t_{1} = {\begin{array}{l} 0 s \\ 2 {(v_{0 x})}_{F} / {(a_{0 x})}_{B} = 8.0 s \end{array} \end{array}

Interestingly, there are two solutions. That’s not surprising, when you think about it, because the line and the parabola of the position graphs have two intersection points: when Fred first passes Betty, and 8.0 s later when Betty passes Fred. We’re interested in only the second of these points. We can now use either of the distance equations to find ${(x_{1})}_{B} = {(x_{1})}_{F} = 160 m$ . Betty has to drive 160 m to overtake Fred.

REVIEW $160 m \approx 160 yards$ . Because Betty starts from rest while Fred is moving at $20 m/s \approx 40 mph,$ needing 160 yards to catch him seems reasonable.

NOTE The purpose of the Review step is not to prove that an answer must be right but to rule out answers that, with a little thought, are clearly wrong.

2.4.3 The Constant-Acceleration Model

MODEL 2.2 Constant acceleration

PROBLEM-SOLVING STRATEGY 2.1
Kinematics with constant acceleration

EXAMPLE 2.10 The motion of a rocket sled

FIGURE 2.23 Pictorial representation of the rocket sled.

EXAMPLE 2.11 A two-car race

FIGURE 2.24 Pictorial representation for Example 2.11.