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Acceleration

The time rate of change of velocity.

MathematicalRepresentation"> Mathematical Representation

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One-DimensionalAcceleration"> One-Dimensional Acceleration

UtilityoftheOne-DimensionalCase"> Utility of the One-Dimensional Case

As with all vector equations, the equations of kinematics are usually approached by separation into components. In this fashion, the equations become three simultaneous one-dimensional equations. Thus, the consideration of motion in one dimension with acceleration can be generalized to the three-dimensional case.

GraphicalRepresentationsofAcceleration"> Graphical Representations of Acceleration

The one-dimensional form of the definition of acceleration:

indicates that acceleration is the slope of a velocity versus time graph.

Since velocity is the derivative of position:

we can also write:

Thus, we can also see the effects of acceleration in a position versus time graph like that shown here:

Consider the position vs. time graph shown above. If you were to lay a ruler along the curve of the graph at the origin, the ruler would have to be horizontal to follow the curve, indicating zero slope. Thus, the velocity is zero at the origin. As you follow the curve, however, the ruler would have to be held at a steeper and steeper angle (see the lines added in the graph below). The slope grows with time, indicating that the velocity is becoming more and more positive (the speed is increasing). This positive change in velocity indicates a positive acceleration. In calculus terminology, we would say that a graph which is "concave up" or has positive curvature indicates a positive acceleration.

Examples that use the graphical representations of acceleration are:

TheDangerofDeceleration"> The Danger of Deceleration

Acceleration poses a problem for the specialized vocabulary of physics. The other two major kinematical quantities, velocity and position, have a related scalar quantity. For instance, distance is a scalar that is related to displacement. Speed is a scalar that is related to velocity. If we are discussing instantaneous velocity, then speed is the magnitude of velocity. The acceleration can also be discussed in terms of a vector acceleration or simply the magnitude, but for acceleration we have no special term for the magnitude. The vector is called "the acceleration" and the magnitude is "the magnitude of the acceleration". This can result in confusion.

This problem is exacerbated by the fact that in everyday language, we often discuss distance, speed and acceleration. The everyday definitions of distance and speed are basically equivalent to their physics definitions, since we rarely discuss direction of travel in everyday speech and these quantities are scalars in physics (no direction). Unfortunately, in physics, we usually use the term "acceleration" to refer to a vector, while in everyday speech it denotes a magnitude.

The difficulties do not end there. Everyday usage does make one concession to the vector nature of motion. When we discuss acceleration in everyday speech, we usually specify whether the object is "accelerating" (speeding up) or "decelerating" (slowing down). Both terms imply a change in velocity, and so in physics we can call either case "accelerating". The physics way of explaining the difference is:

everyday term	physics equivalent
acceleration	acceleration and velocity point in the same direction
deceleration	acceleration points in the direction opposite the velocity

To understand the physics definition, imagine a child on a playground swing. If you want to help the child swing faster, you must push them in the same direction as they are currently moving (so the acceleration of your push is in the same direction as the child's velocity). If you want to help them slow down, you must push them in the direction opposite their current motion (so that the acceleration of the push points opposite to the velocity).

The difference between acceleration and deceleration (in the everyday sense) can also be illustrated graphically:

positive acceleration positive velocity "accelerating"	negative acceleration negative velocity "accelerating"	negative acceleration positive velocity "decelerating"	positive acceleration negative velocity "decelerating"

Both the graphs that show "acceleration" have slopes that are steepening with time. The only difference is that one of the graphs has a steepening positive slope and the other has a steepening _negative slope. Both graphs showing "deceleration" have slopes that are approaching zero as time evolves. (Again, one has a negative slope and one has a positive slope.)

ConstantAcceleration"> Constant Acceleration

Integration with Respect to Time "> Integration with Respect to Time

If acceleration is constant, the definition of acceleration can be integrated:

For the special case of constant acceleration, the integral yields:

which is equivalent to:

We can now substitute into this equation the definition of velocity,

which gives:

We can now integrate again:

to find:

We finish up with some algebra:

which is equivalent to:

Integration with Respect to Position "> Integration with Respect to Position

The definition of acceleration can also be integrated with respect to position, if we use a calculus trick that relies on the chain rule. Returning to the definition of acceleration:

we would like to find an expression for v as a function of x instead of t. One way to achieve this is to use the chain rule to write:

We can now elminate t from this expression by using the defnition of velocity to recognize that dx/dt = v. Thus:

which is easily integrated for the case of constant acceleration:

to give:

One-DimensionalMotionwithConstantAcceleration"> One-Dimensional Motion with Constant Acceleration

Four or Five Useful Equations "> Four or Five Useful Equations

For a time interval during which the acceleration is constant, the instantaneous acceleration at any time will always be equal to the average acceleration. Thus, by analogy with the definition of average velocity, we can write:

Taking this equation as a starting point and using the relationship between average velocity and position

lets us derive five very important equations.

A Training Flight (Some Typical Exercises)"> A Training Flight (Some Typical Exercises)

The Piper Tomahawk, widely used for flying lessons, has a liftoff speed of about 55 knots and a landing speed of about 46 knots. The listed ground-roll for takeoff is 820 feet, and for landing is 635 feet.

Part A

Assuming constant acceleration, what is the Tomahawk's acceleration during the takeoff run?

Solution

System:

In each of the parts, we will treat the Tomahawk as a point particle.

Interactions:

We are assuming that the thrust from the plane's engine, air resistance, rolling friction, and all other influences add up to give a constant acceleration.

Model:

One-Dimensional Motion with Constant Acceleration.

Approach:

Once we have determined that we are using 1-D Motion with Constant Acceleration, we have essentially reduced the problem to math. The Constant Acceleration model is somewhat unique in that there is an overabundance of equations (Laws of Change) to choose from. In order to select the proper equation or equations, we have to clearly understand the information presented in the problem (often called the givens) and also what we are asked to find (the unknowns).

To understand our givens, it is a good idea to first develop a coordinate system. By drawing out the runway and marking where the plane starts its takeoff run (x = 0 feet in the picture) and where it lifts off the ground (820 feet in the picture) we can see that the 820 feet given in the problem statement is a distance not a position. The difference between the plane's liftoff position and the starting point is 820 feet. Thus, in terms of the equations, we have both x (= 820 feet in our coordinates) and x_i (= 0 feet in our coordinates).

We also know that v, the speed at liftoff is 55 knots. Finally, we assume that the starting speed, v_i = 0 knots based upon the setup of the problem. Thus, our givens are:

givens

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[ x_

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= \mbox

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] [ x = \mbox

Unknown macro: {820 feet}

] [v_

= \mbox

Unknown macro: {0 knots}

] [ v = \mbox

Unknown macro: {55 knots}

]\end

The unknown for this problem is the acceleration, since that is what we are asked to find. We can now look over the many possible Laws of Change to see which will allow us to use our givens to solve for our unknowns. In this case, it is pretty clear that the best choice is:

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^2 + 2a(x - x_

) ]\end

This equation uses all of our givens, and it contains our unknown. The other possibilities that contain the acceleration are ruled out because they also contain time. We have no information about the elapsed time for the takeoff.

Once we have the equation, we can solve symbolically for the unknown using algebra. It is good practice to solve algebraically with variables before substituting numbers. The only exception is that you are encouraged to substitute any zeros before doing the algebra. Substituting zeros simplifies the equation. In this case, because x_i and v_i are each zero, we can write:

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which greatly simplifies the algebra needed to isolate the acceleration. We find:

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Now, we would like to substitute in the given values, but we have an important problem. The units of our givens are mismatched. We have x in units of feet, and v in units of knots. If we substitute into the equation we have found, we will recover an acceleration in the units of square knots per foot. This is certainly not a standard unit of acceleration. Because the equations of motion with constant acceleration involve many quantities with different dimensions, it is good practice always to convert to [SI units]. The appropriate [conversions] in this case are: 1 foot = 3.281 m and 1 knot = 0.514 m/s. After conversions, our (nonzero) givens are:

converted givens

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[x = \mbox

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] [ v = \mbox

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We have retained three digits for the velocity (and also for the position, though it is not obvious) even though the original given, 55 knots, contained only 2 significant figures. This is not a mistake. When performing conversions, you should treat them as intermediate steps, not final answers. Retain extra digits through all intermediate steps and round at the end. (This is one important way that solving with symbols before substituting numbers will help you. It reduces the number of intermediate calculations and hence limits the opportunities for rounding errors to enter the calculation.)

With our converted givens, we are ready to substitute, finding:

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)^{2}}{2(\mbox

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)} = \mbox

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^

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]\end

This is basically the answer, though the problem is slightly ambiguous. The word "acceleration" could stand for the vector or simply the magnitude. The magnitude is 1.6 m/s ². Reporting the full vector is tricky, since we are not told which way the plane is moving as it takes off. The best we could do is to say that the vector acceleration is 1.6 m/s ² in the direction of the plane's movement.

Part B

What is the elapsed time from the instant the plane begins its takeoff run until it lifts off the ground?

Solution

System, Interactions and Model: The system, interactions and model are the same as in Part A.

Approach:

Using the work we did to convert in Part A, we have the givens:

givens

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= \mbox

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][x = \mbox

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] [v_

= \mbox

Unknown macro: {0 m/s}

] [v = \mbox

Unknown macro: {28.3 m/s}

][ a = \mbox

Unknown macro: {1.6 m/s}

^

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]\end

Note that we have used the acceleration of Part A as a given for Part B. This extra given makes an enormous difference in the next step of the problem. Instead of one path to the solution, we can now find at least three different equations among the model's Laws of Change that will work. This kind of redundancy is common in this particular model. To illustrate, we will solve the problem using different methods. For simplicity, in using each of the methods, we will choose to make t_i = 0 seconds. (Just as we chose x_i = 0 in setting up our coordinate system.)

Unless explicitly told to assume a certain t_i, we will generally take t_i = 0 to simplify the equations.

Method 1

Suppose you were unable to solve Part A. It is still possible to solve Part B without knowing that the acceleration is 1.6 m/s ². The equation:

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becomes, after using the zeros:

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v t ] \end

which is solved to give:

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] \end

Method 2

If you did solve Part A, the simplest approach is to use the simplest of the Laws of Change:

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) = at ] \end

which gives:

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Method 3

You could also choose the slightly more complicated:

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a(t-t_

)^

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= \frac

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a t^

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]\end

giving:

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\begin

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[ t = \pm \sqrt{\frac

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{a}} = \pm \mbox

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] \end

We select the plus sign, since the plane began its run at t_i = 0 s and it cannot possibly lift off before it has begun to move!

Whenever taking a square root in physics it is important to use your understanding of the problem to choose the appropriate sign.

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Part C

How far does the plane travel in the first 9.0 seconds of the takeoff?

Solution

System, Interactions and Model: As in the previous parts.

Approach:

Based upon the results of Part B, we know that the plane has not reached its takeoff speed within 9.0 seconds, and also that it has not traveled its full takeoff distance. We can still use the result of Part A that the acceleration is 1.6 m/s ², however, since we have been assuming the acceleration is constant throughout the run. Thus, we are left with the truncated set:

givens

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= \mbox

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][ v_

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= \mbox

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][t_

= \mbox

Unknown macro: {0 s}

] [t = \mbox

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] [ a = \mbox

Unknown macro: {1.6 m/s}

^

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]\end

The unknown that we seek is x, the location at t = 9.0 s. The most direct approach is to use:

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+ v_

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a(t-t_

)^

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= \frac

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at^

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] \end

No algebra is necessary, since x is already isolated. We can simply substitute to find:

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It is important to note that the plane travels much less than 1/2 the takeoff distance in about half the takeoff time. In fact, based upon the t² dependence of the formula we just employed, we expect that the plane should cover only 1/4 of the takeoff distance in the first half of the run. That leaves it to cover the remaining 3/4 in the second half. The reason for this difference is simple: the plane is constantly accelerating, and so it moves faster during the second half of the run.

Part D

What is the acceleration during landing?

Solution

System, Interactions and Model: As in the previous parts.

Approach:

This problem is essentially identical in structure to Part A, except that when landing, the plane's initial speed is nonzero and its final speed is zero. Thus, we proceed in the same fashion as in Part A. We first define a coordinate system:

and summarize our givens (this time already converted to SI units):

converted givens

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[ x_

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= \mbox

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][x = \mbox

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][ v_

= \mbox

Unknown macro: {23.6 m/s}

][ v = \mbox

Unknown macro: {0 m/s}

]\end

Then, we choose the appropriate equation:

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^

+ 2a(x-x_

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) ] \end

and substitute zeros:

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^

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+ 2ax ] \end

Finally, we solve symbolically and substitute:

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^{2}}

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= -\mbox

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^

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]\end

Again, the question is ambiguous about what is expected. The magnitude of this acceleration is simply 1.4 m/s ². To report it as a vector, the best option is probably to say 1.4 m/s ² in the direction opposite the plane's motion.

Part E

How far does the plane travel in the first 9.0 seconds of the landing?

Solution

System, Interactions and Model: As in previous parts.

Approach:

This problem is similar to Part C. Again, 9.0 seconds is not the full time over which the plane accelerates (can you find the total time needed to bring the plane to rest?) and so we cannot assume that the final speed is zero or the final position is 194 m. The givens that we have, including the result of Part D, are:

converted givens

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= \mbox

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][ v_

= \mbox

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][ a= -\mbox

Unknown macro: {1.4 m/s}

^

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]\end

It is extremely important to retain the negative sign on the acceleration when using it in the equations. We will see why in a moment.

As in Part C, we will choose t_i = 0 s, and we will use the equation:

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+ v_

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a(t-t_

)^

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= v_

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t + \frac

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a t^

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] \end

Note we cannot cancel the initial velocity for the case of a landing plane.

Again, no algebra is required, so we substitute:

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[ x = (\mbox

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)(\mbox

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) + \frac

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(-\mbox

Unknown macro: {1.4 m/s}

^

)(\mbox

)^

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= \mbox

Unknown macro: {156 m}

] \end

To see the importance of the negative sign, calculate the position after 9.0 s using a positive acceleration. What do you find? The negative ensures that the plane's deceleration causes it to cover less distance than it would if it were simply coasting at its initial speed. Similarly, a positive acceleration will cause the plane to cover extra distance.

The Utility of Constant Acceleration"> The Utility of Constant Acceleration

Stringing together a series of constant velocity segments is not usually a realistic description of motion, because real objects cannot change their velocity in a discontinuous manner. This drawback does not apply to constant acceleration, however. Objects can have their acceleration changed almost instantaneously. For example, you could be coasting along in a car at a constant 60 mph with zero acceleration when suddenly you see traffic stopped ahead. If you slam on the brakes, your car will still be going 60 mph for an instant, then it will drop to 59, 58, 57, etc. Your acceleration, on the other hand, has almost instantly changed from zero to a substantial acceleration directed opposite your motion. You can feel this abrupt change as a passenger as you are forced against your seatbelt. Similarly, when an airplane begins its takeoff run, you can feel yourself suddenly pressed back in your seat as the plane's acceleration changes almost instantaneously from 0 to a significant forward acceleration. Because of this, it is often reasonable to approximate a complicated motion by separating it into segments of constant acceleration.

An Exercise in Continuity"> An Exercise in Continuity

Suppose an object is moving along a one-dimensional position axis. The object starts its motion at t = 0 at the position x = 0 and with velocity v = 0. It has an acceleration of +2.0 m/s². After 4.0 seconds, the object's acceleration instantaneously changes to - 2.0 m/s². Plot velocity and position versus time graphs for the first 8.0 seconds of the object's motion.

Solution

System:

The object will be treated as a point particle.

Interactions:

The accelerations as described in the problem.

Model:

One-Dimensional Motion with Constant Acceleration, applied twice to cope with the change in acceleration.

Approach:

Understand the Limitations of the Model

We will have to use the model twice. Even though the acceleration has a size of 2.0 m/s² throughout the motion, the direction of the acceleration changes at t = 4.0 s. Therefore, the acceleration is not constant over the first 8 seconds of the motion. The acceleration is, however, separately constant in the time intervals from 0 to 4 seconds and from 4 to 8 seconds. We will therefore have to use the model twice: once to describe the motion from t = 0 seconds to t = 4.0 seconds, and once more to describe the motion from t = 4.0 seconds to t = 8.0 seconds.

The First Interval

For the first part of the motion, our givens are:

We begin by finding the velocity. The simplest Law of Change appropriate to our givens is:

which, after substituting the givens, tells us:

This is the equation for a line with slope 2 and intercept 0, giving the graph:

We then find the position. The most direct Law of Change is:

which yields the parabolic graph:

The Second Interval

We now wish to analyze the second part of the motion. In this part of the motion, we must change our givens. We now have t_i = 4.0 s, t = 8.0 s, and a = – 2.0 m/s². Unfortunately, this is not enough. We do not know x_i, x, v_i or v. We have too few givens to proceed.

The answer to this dilemma is simple. We have just derived expressions that give x and v for any time between 0 s and 4.0 s. We would like to know x and v at 4.0 s. Thus, we can use the final time of the first part of the problem to obtain the initial conditions for the second part. From our graphs or from the equations, we can complete our list of givens for the second part:

This allows us to write an equation for the velocity:

and the position:

Putting them Together

These new expressions can be used to finish the plots we started in the first part. Plotting the functions from the first part in the interval t = 0 s to t = 4.0 s and the functions from the second part in the interval t = 4.0 s to t = 8.0 s gives the complete graphs shown below. For completeness, we also show the acceleration graph.

Here we can see that although the acceleration is discontinuous (has a break in the graph at t = 4.0 s) the velocity is not discontinuous. The velocity is a connected graph, though it does display a kink at 4.0 s, when the graph suddenly takes a sharp bend. This kink is a sign of a discontinuous 1st derivative (the 1st derivative of velocity with respect to time is acceleration, which is discontinuous). The position, however, is completely smooth. It is connected and displays no kinks. (The first derivative of position is velocity, which is continuous.)

Example Problem – Fan-Powered Ice Boat"> Example Problem – Fan-Powered Ice Boat

Pictured here is the University of Manitoba Center for Earth Observation Science's air/ice boat "Skippy" (photo courtesy CEOS).  Skippy has a large fan at the back to allow it to accelerate as it slides across icy surfaces.  Suppose you are piloting a similar craft on very slippery ice which will not slow the boat at all if it is coasting.  Suppose also that the fan on your boat can be reversed instantanously, switching its direction of thrust from forward to backward.  Assume the action of the fan (when it is on) always produces an acceleration with the same constant magnitude.  Air resistance is negligible.

For the questions in this example, use the following coordinate system (illustrated above).  You have a Base Camp at position x=0.  Your assignment is to make observations of ice thickness and atmospheric conditions at two stations.  Station One is 1 mile east of Base Camp, and Station Two is 2 miles west of Base Camp.  Take east to be the positive x-direction. 

Problem 1

Part A

Suppose you have left Base Camp and are halfway to Station One.  You have been accelerating to the east the entire trip, but you now realize you have forgotten your gloves.  You immediately flip the fan control switch to backward, reversing the direction of the thrust.  Describe in words what will happen to your position and your velocity from the instant you reversed the fan. 

Solution

System:

The boat and its contents will be treated as a point particle.

Interactions:

An external force from the action of the fan (actually from the air that the fan is pushing).

Model:

One-Dimensional Motion with Constant Acceleration

Approach:

Once the fan has been reversed, the acceleration is in the opposite direction from the velocity. Thus, the boat will immediately begin to slow down. As it slows, it will continue to move east. In the language of our one-dimensional system, the velocity immediately begins to decrease with time, but the position continues to increase until the instant that the velocity has decreased to zero.

It is very tempting to assume that as soon as the fan is reversed, the boat begins to move backward. This is not the case, however. Consider the case of a commercial jet landing at over 100 mph. The pilots will usually reverse the engines within a few seconds of landing to help slow the plane. If the plane were to immediately reverse direction, the effect on the passengers would be as if it had hit a brick wall and bounced off! Instead, the plane continues along the runway while gently decreasing its speed.

One of the (many) reasons that your intuition might tell you that reversing the engine immediately reverses the direction is experience operating automobiles. In a car, the transmission is not designed to allow you to reverse the engine to slow the car. When the engine is in reverse either the car is moving backward or you are about to spend a few thousand dollars on your transmission. What feature do cars have (that is not present on the air/ice boat) to compensate for the fact that the engine cannot produce a significant acceleration in the direction opposite the car's motion?

The motion is not finished when the boat stops, however. If the fan is left in reverse, the continued thrust will now begin to accelerate the boat backward. In one-dimensional language, the velocity reaches zero and continues to decrease, so the speed (absolute value of velocity) is now increasing. The boat will therefore begin to move west, and with increasing speed. Thus, the position vs. time graph will begin to decrease with a steepening slope.

It is vitally important to remember that an object cannot remain at rest for more than an instant if it is acted upon by a constant acceleration.

Part B

Sketch rough graphs of your velocity and position as a function of time from the instant the fan was reversed.

Solution

System, Interactions and Model: As in Part A.

Approach:

We construct the graphs that illustrate our answers to Part A.

For sketched graphs, you needn't put numbers on the axes, even though we did that for the position graph here. It is important, however, that you make your time axes consistent with each other. In this case, at the second division of the time axis, the velocity goes to zero. Thus, on the position graph we expect to see the position's slope at zero (which it is). Graders will always check for this kind of consistency.

Part C

  Given that the boat was halfway to Station One before you reversed it, that it started from rest at Base Camp, and that the fan was accelerating the boat forward until the instant you reversed it, where will the boat be when it (instantaneously) comes to rest after you have reversed the fan? 

System and Interactions: As in Part A.

Model:

To correctly answer this part, you will have to model the boat's motion starting from Base Camp. The model is still One-Dimensional Motion With Constant Acceleration, but it has to be applied twice. First, there is the eastward acceleration from the time that you leave Base Camp until you reverse the fan. Then, there is the westward acceleration model of Part A.

Approach:

This time we are asked for a quantitative answer. You can solve this with equations, but you would have to make up numbers. A better way to do this is to make use of the graphs as a tool to help you understand the symmetry of the equations. Here is an extension backward in time of the velocity graph of Part B which shows the acceleration from the forward thrust of the fan as you started out from Base Camp. We know that the slope should simply be the opposite of the slope when the fan is reversed. Further, we assume that you started at rest. Thus, the red part of the graph is a reasonable representation of your motion with the fan accelerating you toward Station One. Notice the obvious symmetry with the part of the graph from Part B that has been colored blue in the new graph. It is clear that the area under these lines is the same. Thus, using what we know about the relationship between velocity and position, we conclude that the distances traveled are the same. Since the red part corresponds to traveling halfway to Station One, the blue part corresponds to the same distance. The boat has made it to Station One just as it comes to rest (note that the blue part terminates at zero velocity).

Part D

Suppose that once you have reversed the fan, you keep the fan reversed and on.  Will you arrive safely back at Base Camp?

Solution

System, Interactions and Model: As in the previous parts.

Approach:

You will certainly reach Base Camp, but it is perhaps not reasonable to say you have "arrived safely". You might worry, since by the time you reach Base Camp you will be traveling fast enough that it would take you a mile to stop if you had to rely only on the reverse thrust of the fan. Unless you can find some more efficient (but safe) way of stopping the boat, you will have to repeat the tedious process of turning the boat around again if you want to get your gloves.

Problem 2

Suppose you are making your daily rounds.  This involves: 

1. starting from rest at the Base Camp
2. traveling to Station Two
3. remaining at Station Two for the same amount of time it took to get there
4. turning the boat around and traveling to Station One (without stopping at Base Camp)
5. remaining at Station One for the same amount of time you were at Station Two
6. turning the boat around and returning to Base Camp and stopping there

Part A

Sketch graphs of your position and velocity as a function of time for your entire daily rounds.  Assume that whenever you travel, you accelerate toward the destination for exactly half the trip and then decelerate for the other half.

System, Interactions and Model: As in the previous problem.

Answer:

Note that the peaks of the velocity (solid dark red graph) do not occur at the same time as the position (dotted blue graph) peaks! Rather, they correspond to changes in the curvature of the position graph.

Part B

Divide your graphs into segments and label each segment with the position of the fan switch (forward, backward or off) during that segment. Remember that in your daily rounds, you always turn the boat before setting off from Base Camp or the Stations to point in the direction you are moving.

Solution

System, Interactions and Model: As in the previous problem.

Approach:

In the plots shown here, green lines mean the switch is "forward", red lines mean "backward" and blue means "off".

Problem 3 - Challenge

In Problem 2 we assumed that the best way to travel is to speed up for half the trip and then slow down for the second half.  This gives your craft the greatest controllable speed (assuming you don't want to hit something at your destination to stop), but it also requires you to decelerate much earlier than if you just turned off the fan and coasted for a while at some intermediate speed.  Assuming you want to minimize the time to your destination, what is the best way to use the fan?

Near-Earth Gravity"> Near-Earth Gravity

One of the most important applications of One-Dimensional Motion with Constant Acceleration is describing the motion of objects moving vertically under the influence of [gravity]. For objects near the surface of the earth, gravity creates a constant pull directed essentially straight toward the ground.

Interestingly, the gravitational acceleration produced by the earth is not only constant on objects near its surface, but it is also exactly the same magnitude for all objects if air resistance is negligible. That value is:

Example Problem – Keys Please"> Example Problem – Keys Please

A college student with a third floor dorm room is exiting the dorm when he suddenly realizes he has forgotten his keys. Rather than run back to the room, he calls his roommate and goes to stand under the dorm room window. The roommate brings the keys to the window.

Part B

Suppose that, instead of dropping the keys, the roommate tosses them straight up with an initial speed of 3.3 m/s. The keys are released from a height h= 5.0 m above the outstretched hand of the forgetful student. How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

Solution

System, Interactions and Model: The system, interactions and model are the same as for Part A.

Approach:

We will illustrate two separate but completely equivalent ways to do this problem. The first way is faster, but requires familiarity with the quadratic equation. The second way avoids the quadratic equation by making use of symmetry, but it requires more physical insight.

With Quadratic

The problem is identical to Part 1 except that v_i is not zero. Thus, the equation that we have to solve is:

This is a quadratic equation. It is a good idea to rearrange it:

so that it is clear that we have an equation of the form

if we choose:

Using these assignments in the quadratic equation gives:

To obtain a positive time, the positive sign must be chosen since the radical expression will clearly evaluate to be larger than v_i, so:

Without Quadratic

Freefall (and later, projectile) problems can often be usefully broken into two parts and analyzed in a mathematically straightforward fashion. The point at which we will separate the problem into two parts is the point of maximum height. As we will see, this is a useful choice because the velocity goes to zero at that point. We begin by analyzing the upward motion of the keys.

Considering only the trip up to maximum height (shown in blue in the figure at right), we know both the initial and final velocities. Since we also have the acceleration (gravity), this information allows us to use the simplest available Law of Change in our model:

so, using a=-g and v=0:

We have the time, which is what we wanted, but we must now make a slight detour and find how far the keys travel during their upward motion. The reason for this is that (as you will see) we will need that height to find the time the keys take to fall down from the peak to the student's hand. Luckily, this distance is easy to find. There are many equations that can be solved for it, but we will choose the one that is mathematically simplest:

Now if we use our expression for t_up and the fact that v = 0 at the peak, we find that the distance traveled up from the roommate's hand is:

.

We are now ready to solve for the time to fall down from the peak (the part of the path shown in red in the figure at right). The solution proceeds exactly as in Part 1, so we use that result. The only change is that the keys are now falling from a total height of:

Then, using the answer to Part A, the time to fall is:

The total time is:

Part C

As a final illustration, suppose that the roommate throws the keys downward with an initial speed of 3.3 m/s. The keys are released from a height h= 5.0 m above the outstretched hand of the forgetful student. How long are the keys in the air from the time the roommate releases them until the instant they land in the student's hand?

Solution

System, Interactions and Model: As in the previous parts.

Approach:

As in Part B, we will illustrate two separate but completely equivalent ways to do this problem. The first way is one step, but requires familiarity with the quadratic equation. The second way avoids the quadratic equation, but it requires two steps and stronger physical insight.

With Quadratic

The solution is exactly the same as for Part B. The only difference is that v_i = -3.3 m/s instead of +3.3 m/s. Thus, we find:

Without Quadratic

This time, the motion has no peak point at which the keys are turning around. Thus, we cannot break up the motion as we did in Part B. Instead, we use a two step process. It is very valuable in freefall (and projectile) problems to know the initial and final velocities. In this case, we only have the initial velocity as a given, but we can find the final velocity as a first step toward solving the problem. To do this, we must recognize that without time the only Law of Change for our model that can be solved for the final velocity is:

Using the familiar substitutions, this becomes:

In this case, the minus sign must be chosen, since the keys are clearly moving downward just before they are caught.

Now that we have both the initial and final velocity, we can use the very simple Law of Change:

to find:

which gives:

Part D – Challenge

Can you show that the answer to Part C can be obtained from the answer to Part B by subtracting the time the keys spent above 5.0 m?

Catch-Up Problems By Example"> Catch-Up Problems By Example

Example Problem – Speed Trap"> Example Problem – Speed Trap

A common type of kinematics problem involves one person or object catching up to another person or object. Here we illustrate a few typical variations of this problem.