Equations of Motion

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© 1998-2008 by Glenn Elert -- A Work in Progress
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Discussion

In order to be accurate, the title of this section should be "One Dimensional Equations of Motion for Constant Acceleration". Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. The equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.

Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time -- not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute metaphysical certainty.

So what good is this section then? Well, in many instances, it is useful to assume that an object did or will travel along a path that is essentially straight and with an acceleration that is nearly constant. That is, any deviation from the ideal motion can be essentially ignored. Motion along a curved path may also be effectively one-dimensional if there is only one degree of freedom for the objects involved. A road might twist and turn and explore all sorts of directions, but the cars driving on it have only one degree of freedom -- the freedom to drive in one direction or the opposite direction. (You can't drive diagonally on a road and hope to stay on it for very long.) In this regard, it is not unlike motion restricted to a straight line. Approximating real situations with models based on ideal situations is not considered cheating. This is the way things get done in physics. It is such a useful technique that we will use it over and over again.

Our goal in this section, is to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity, displacement, and time. There are three ways to pair them up: velocity-time, displacement-time, and velocity-displacement. In this order, they are also often called the first, second, and third equations of motion, but there is no compelling reason to learn these names. Since we are dealing with motion in a straight line, the symbol x will be used for displacement. Direction will be indicated by the sign (positive quantities point in +x direction, while negative quantities point in the −x direction). Determining which direction is positive and which negative is entirely arbitrary. The laws of physics are isotropic; that is, they are independent of the orientation of the coordinate system. As long as you are consistent, it doesn't matter. Some problems are easier to understand and solve, however, when one direction is chosen positive over another.

velocity-time

The relation between velocity and time is a simple one during constantly accelerated, straight-line motion. Constant acceleration implies a uniform rate of change in the velocity. The longer the acceleration, the greater the change in velocity. If after a time velocity increases by a certain amount, after twice that time it should increase by twice that amount. Change in velocity is directly proportional to time when acceleration is constant. If an object already started with a certain velocity, then its new velocity would be the old velocity plus this change. You ought to be able to see the equation in your mind's eye already.

This is the easiest of the three equations to derive formally. Start from the definition of acceleration, expand the Δv term, and solve for v as a function of t.

a =  Δv  =  v − v0  
Δt Δt
         
  v =  v0 + aΔt   [1]

Since acceleration is also the first derivative of velocity with respect to time, this equation can also be derived using calculus. Just reverse the action of the definition. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. Since acceleration is assumed constant, this is quite easy.

    a =  dv  
dt
    dv =  a dt
  v   Δt    
      ⌠
⌡		  dv =  ⌠
⌡		  a dt  
  v0   0    
  v − v0 =  aΔt  
      v =  v0 + aΔt   [1]

The symbol v0 (v nought) is called the initial velocity. It is often thought of as the "first velocity" but this is a rather naïve way to describe it. Take the case of a meteor hurtling towards the earth. What is its initial velocity? If you want v0 to be the first velocity, then the problem is over before it started. Who could possibly say what a meteor's velocity was at its birth? There is no way to answer this question. A better definition would be to say that an initial velocity is the velocity that a moving object has when it first becomes important in a problem. Say the meteor was spotted deep in space and the problem was to determine its trajectory, then the initial velocity would be the velocity at the time it was observed. But if the problem were to determine its velocity on impact, then it's initial velocity would more likely be the velocity it had when it entered the earth's atmosphere. In this case, the answer to, "What's the initial velocity?" is "It depends". This turns out to be the answer to a lot of questions.

The symbol v is then the velocity some time Δt after the initial velocity. It is often called the final velocity but this does not make it an object's "last velocity". Take the case of the meteor. What velocity is represented by the symbol v? If you've been paying attention, then you should have anticipated the answer. It depends. It could be the velocity of the meteor as it passes by the moon, as it enters the earth's atmosphere, or as it strikes the earth's surface. It could also be the meteorite's velocity as it sits in the bottom of a crater (zero). Is this then the final velocity? Who knows. Someone could extract the meteorite from its hole in the ground and drive away with it. Is this relevant? Well, maybe. It depends. There's no rule for this kind of thing. You'll have to understand the problem and then make a decision for yourself.

The last part of this equation aΔt is the change in the velocity from the initial value. Recall that a is the rate of change of velocity and that Δt is the time interval since the object had its initial velocity v0. Rate multiplied by time equals change. Thus if an object were accelerating at 10 m/s2, after 5 s it would be moving 50 m/s faster than it was initially. If it started with a velocity of 15 m/s, its velocity after 5 s of acceleration would be 65 m/s.

displacement-time

The displacement of a moving object is directly proportional to both velocity and time. Move faster. Go farther. Move longer (as in longer time). Go farther. Acceleration compounds this simple situation. Now the velocity is also directly proportional to time. Try saying this in words and it sounds ridiculous. "Displacement is directly proportional to time and directly proportional to velocity, which is directly proportional to time." Time is a factor twice, making displacement proportional to the square of time. A car accelerating for two seconds would cover four times the distance of a car accelerating for only one second (22 = 4). A car accelerating for three seconds would cover nine times the distance (32 = 9).

Would that it should be so simple. This example only works when initial velocity is zero. Change in displacement is proportional to the square of time when acceleration is constant and initial velocity is zero. A true general statement would have to take into account any initial velocity and how the velocity was changing. This results in a terribly messy proportionality statement. Change in displacement is directly proportional to time and proportional to the square of time when acceleration is constant. A function that is both linear and square is said to be quadratic, which allows us to compact the previous statement considerably. Change in displacement is a quadratic function of time when acceleration is constant

Proportionality statements are useful, but not as concise as equations. We still don't know what the constants of proportionality are for this problem. The only way to answer that is through algebra.

Start with the definition of velocity, expand Δx, and solve it for displacement.

v =  Δx    
Δt
Δx =  vΔt    
  x − x0 =  vΔt      
  x =  x0 + vΔt   [a]

To continue, we need to resort to a little trick first published in the Fourteenth Century at Merton College, Oxford (and sometimes called the Merton Rule). When acceleration is constant, the velocity will change uniformly from its initial value to its final value and the average will lie halfway between the extremes. Thus, the average velocity is just the arithmetic mean of the initial and final velocities. Average velocity is the average of the final and initial velocities when acceleration is constant.

v =  v + v0
2
[4]

Substitute the first equation of motion [1] into this equation (4) and simplify with the intent of eliminating v.

v =  v + v0  =  (v0 + aΔt) + v0  =  2v0 + aΔt   = v0 + ½ aΔt [b]
2 2 2

Finally, substitute [b] into [a] and solve for x as a function of t.

  x =  x0 + vΔt    
  x =  x0 + (v0 + aΔt)Δt    
  x =  x0 + v0Δt + ½ aΔt2   [2]

Since velocity is also the first derivative of displacement with respect to time, this equation can also be derived using calculus. In fact, it's much easier than using algebra. Just reverse the action of the definition. Instead of differentiating displacement to find velocity, integrate velocity [1] to find displacement.

v =  dx        
dt
dx =  v dt      
  x Δt   Δt    
      ⌠
⌡		  dx =  ⌠
⌡		  v dt =  ⌠
⌡		  (v0 + aΔt)dt  
  x0   0   0    
  x − x0 =  v0Δt + ½ aΔt2  
  x =  x0 + v0Δt + ½ aΔt2   [2]

The symbol x0 (ex nought) is the initial displacement. Many times, this value is zero and if it isn't, we can make it so. If you ask me, "When should we do this?" I would say, "It depends on the problem," and leave it to you to decide. There is no rule that you can memorize in this case. You have to understand what the equation says and then learn how to apply it to a particular situation. Similarly x is often called the final displacement, but this does not make it the "last displacement", rather it is the displacement at the end of the time interval during which the acceleration was constant.

Something else to notice is the similarity between equations [2] and [a]. When acceleration is zero, our second equation of motion reverts to a rearranged constant velocity equation. As predicted, displacement is in part directly proportional to time and in part directly proportional to time squared.

  x =  x0 + v0Δt + ½ aΔt2   [2]
  x =  x0 + vΔt   [a]

Although the velocity symbols in the two equations may look different, they do indeed represent the same quantity. If there is no acceleration, then the velocity is constant, which means that the initial velocity is the same as the final velocity is the same as the average velocity. The acceleration term at the end is an adjustment to the constant velocity equation to account for the the fact that the velocity is changing. A positive acceleration would increase the displacement and a negative acceleration would decrease it. This is exactly what one would expect. If an object's velocity was increasing, it would move farther than if it had stayed at a constant velocity. Likewise, if an object's velocity was decreasing, it would have a smaller displacement than if its velocity were constant. It's good to see that the equations behave in a realistic manner. Otherwise all this math would be a waste of time.

velocity-displacement

We have just seen that velocity is directly proportional to time and displacement is proportional to time squared. With a little bit of thinking, a new proportionality statement should be apparent. Change in displacement is proportional to the change in the square of velocity when acceleration is constant. This statement is particularly important for driving safety. When you double the speed of a car, it takes four times the distance to stop it. Triple the speed and you'll need nine times the distance. Like the previous relationships, it also depends on initial velocity. Unfortunately determining the effect using reasoning alone is a real chore. Do the algebra instead.

The last two equations each described one kinematic variable as a function of time. It would be nice if we also had an equation that was independent of time. That is, we want to answer the question, "What is the relationship between velocity and displacement?" The method of doing this should be readily apparent. We've got to combine our first two equations of motion together in a manner that will eliminate time as a variable. The easiest way to do that is to solve one equation for time and then substitute it into the other. The second equation of motion is a quadratic and solving it for time would introduce a lot of nastiness into the algebra. It should be apparent that solving the first equation of motion [1] for time and substituting it into the second [2] will be the easier process.

  x =  x0 + v0Δt + ½ aΔt2 [1]     v =  v0 + aΔt [2]  
  x − x0 =  v0Δt + ½ aΔt2   ⇐ Δt =  v − v0  
a
x − x0 =  v0  ⎛
⎝		 v − v0 ⎞
⎠		  +  1  a  ⎛
⎝		 v − v0 ⎞2
⎠
a 2 a
x − x0 =  vv0 − v02  +  v2 − 2vv0 + v02  
a 2a
  2a(x − x0) =  (2vv0 − 2v02) + (v2 − 2vv0 + v02)  
 
    2a(x − x0) =  v2 − v02    
 
  v2 =  v02 + 2a(x − x0)   [3]

As expected, displacement is proportional to velocity squared. (Initial velocity squared is just a constant that must be dealt with.) Unlike the first and second equations of motion, there is no advantage in using calculus to derive the third equation of motion. Algebra is the way to go.

Summary


The One Dimensional Equations of Motion for Constant Acceleration
traditional name equation relationship
1st equation   v = v0 + at   velocity –  time
2nd equation   x = x0 + v0t + ½ at2   displacement –  time
3rd equation   v2 = v02 + 2a(x − x0)   velocity –  displacement
merton rule   v = ½ (v + v0)   average velocity

Problems

practice

  1. Cars cruise down an expressway at 25 m/s. Engineers want to design an interchange for a deceleration of −2.0 m/s2 over 3.0 s.
    1. What velocity will cars have at the end of the approach?
    2. What minimum approach length will satisfy these requirements?
    3. What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)

    Solutions …

    1. What velocity will cars have at the end of the approach?
             
        v0 =  25 m/s     v =  v0 + at  
        a =  −2.0 m/s2       v =  (25 m/s) + (−2.0 m/s2)(3.0 s)    
        Δt =  3.0 s     v =  19 m/s  
        v =  ??      
             
    2. What minimum approach length will satisfy these requirements?
             
        v0 =  25 m/s     Δx =  v0t + ½ at2  
        a =  −2.0 m/s2     Δx =  (25 m/s)(3.0 s) + ½ (−2.0 m/s2)(3.0 s)2  
        Δt =  3.0 s   Δx =  66 m  
        Δx =  ??      
             
    3. What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)
             
        v =  19 m/s     v2 =  v02 + 2aΔx  
        a =  −8.0 m/s2     v02 =  v2 − 2aΔx  
          Δx =  66 m       v02 =  (19 m/s)2 + 2(−8.0 m/s2)(66 m)  
        v0 =  ??     v0 =  38 m/s  
             
  2. A car with an initial velocity of 60 mph needs 144 feet to come to a complete stop. Determine the stopping distance of this same car, under identical conditions with an initial velocity of 30 mph, 20 mph, and 10 mph.

    First method …

    The hard way to solve this problem is to do it the way that many students think is the easy way. That is, the "plug and chug" method shown in the previous examples. This method appears easy since it requires very little thought, but it turns out to be quite demanding.

    First, convert to SI units.

    60 mile   1609 m   1 hour  = 26.8  m   144 feet   1 mile   1609 m  = 43.9 m  
    1 hour 1 mile 3600 s s 1 5280 feet 1 mile
    30 mile   1609 m   1 hour  = 13.4  m  
    1 hour 1 mile 3600 s s
    20 mile   1609 m   1 hour  = 8.94  m
    1 hour 1 mile 3600 s s
    10 mile   1609 m   1 hour  = 4.47  m
    1 hour 1 mile 3600 s s

    Then, calculate acceleration from the data for 60 mph.

      v0 =  26.8 m/s       v2 =  v02 + 2aΔx    
      v =  0 m/s     a =  v2 − v02
      Δx =  43.9 m   2Δx
      a =  ??     a =  −(26.8 m/s)2
        2(43.9 m)
          a =  −8.18 m/s2  

    Now, calculate the distances for the other speeds.

      v2 =  v02 + 2aΔx       Δx =  −(13.4 m/s)2  = 11.0 m    
    2(−8.18 m/s2)
      Δx =  v2 − v02  =  − v02     Δx =  −(8.94 m/s)2  = 4.89 m  
    2a 2a 2(−8.18 m/s2)
        Δx =  −(4.47 m/s)2  = 1.22 m  
    2(−8.18 m/s2)

    And finally, convert back into English units.

    11.0 m   1 mile   5280 feet  = 36 feet
    1 1609 m 1 mile
    4.89 m   1 mile   5280 feet  = 16 feet
    1 1609 m 1 mile
    1.22 m   1 mile   5280 feet  = 4 feet
    1 1609 m 1 mile

    Second method …

    Standard problem solving techniques work, but they're a monumental waste of time in this case. Any small error would destroy the answers and waste personal mental energy as well -- something we all like to avoid. The easy way to solve this problem does not involve any trickery. It requires that you identify and understand the key concepts needed to solve the problem. Halfway through the mass of equations, an important assumption was made. It was assumed that the braking acceleration of the car would remain constant for all initial velocities. This problem is then one of determining the relationship between displacement and velocity. The equation that does this is

    v2 = v02 + 2a(x − x0)

    which shows that displacement is proportional to velocity squared when acceleration is constant and either the the initial or final velocity is zero.

    x ∝ v2

    In this problem we're comparing stopping distances at 30, 20, and 10 mph to those at 60 mph. The square of the ratios of the new velocity to the original velocity will be the ratio of the new stopping distances to the original stopping distance.

    The Easier Way to Solve This Problem
    ratio of new velocity
    to original velocity    		 ratio of new distance
    to original distance    		 new stopping distance
    30 mph∕60 mph = 1∕2 (1∕2)2 = 1∕4 (1∕4) 144 feet = 36 feet
    20 mph∕60 mph = 1∕3 (1∕3)2 = 1∕9 (1∕9) 144 feet = 16 feet
    10 mph∕60 mph = 1∕6 (1∕6)2 = 1∕36 (1∕36) 144 feet = 4 feet

    These are exactly the answers we got after doing all the brainless "plug and chug" calculating of the hard method. Isn't it wonderful to have a brain.

  3. A typical commercial jet airliner needs to reach a speed of 180 knots before it can take off. (A knot is a nautical mile per hour and is very nearly equal to half a meter per second.) If such a plane spends 30 s on the runway estimate …
    1. its acceleration.
    2. the minimum runway length.

    Solutions …

    1. To determine acceleration, I recommend using the definition of acceleration.
                   
      v0 =  0 m/s a =  Δv  =  90 m/s  
      v =  1180 kts ≈ 90 m/s Δt 30 s
      t =  30 s a =  3 m/s2 ≈ ⅓ g
      a =  ??          
                   
    2. There are two ways to determine the runway length. Either method yields the same solution.
      • Using the distance-time equation.
               
        v0 =  0 m/s s =  s0 + v0t + ½ at2
        v =  1180 kts ≈ 90 m/s s =  ½ (3 m/s2)(30 s)2
        t =  30 s s =  1350 m
        a =  3 m/s2    
        s =  ??    
               
      • Or from the average velocity during uniform acceleration.
               
        v0 =  0 m/s s =  vt = ½ (v + v0)t
        v =  1180 kts ≈ 90 m/s s =  ½ (90 m/s + 0 m/s)(30 m/s)
        t =  30 s s =  1350 m
        s =  ??    
               
  4. Write something completely different.
    • Answer it.

conceptual

  1. On well-engineered highways obstacles like bridges, retaining walls, and guardrails are often protected by large yellow barrels filled with water or sand called impact attenuators. Should a wayward car find itself on on a collision course with a relatively immoveable obstacle, an impact attenuator would help reduce the severity of the crash. If one of these barrels provided sufficient protection for an impact at 30 km/h, how many barrels would provide sufficient protection for an impact at …
    1. 60 km/h?
    2. 90 km/h?
    3. 120 km/h?
  2. In a series of tests conducted by the Insurance Institute for Highway Safety (IIHS), a passenger sedan traveling 55 mph came to a stop in 133 feet while a sleeper cab tractor with a loaded trailer required 196 feet. (Answer the following questions without considering the units.)
    1. How do the braking accelerations of these vehicles compare?
    2. How do the braking times of these vehicles compare?

numerical

  1. Determine the minimum thickness of a fully inflated air bag if it must stop a passenger moving at 14 m/s (31 mph) with an acceleration of less than 60 g.
  2. By how much should the front end of a car be designed to crumple such that a front end collision at 30 m/s (67 mph) would not result in an average acceleration greater than 30 g?
  3. At main engine cutoff (MECO), the Space Shuttle is at an altitude of 113 km (70 miles), traveling 7600 m/s (17,000 mph) relative to the earth. This occurs 7 minutes 40 seconds into the mission. Determine the downrange distance of the shuttle from lift off to MECO.
  4. A moving driver not anticipating an accident can apply the brakes fully in about 0.5 s. It takes something on the order of 5 s to stop a car on the freeway once the brakes are locked. Calculate the minimum emergency stopping distance needed at 30 m/s (65 mph).
  5. Researchers at MIT are attempting to build robots that mimic the swimming abilities of fish. Two robotic fish have been constructed as of March 2000: Robotuna and Robot Pike. Tuna and Pike are exceptionally proficient swimmers. According to one researcher, "My project here is to build a swimming Pike. The characteristics that I hope to demonstrate are very quick turning and fast acceleration from a stop. In the wild, Pike accelerate at rates from 8-12 g during a start from a standstill to 6 m/s. While we may not be able to achieve these wild numbers, half or a quarter of this acceleration would still demonstrate that flapping foil propulsion is certainly capable of higher accelerations than a propeller."
    1. How long does it take a wild pike to reach its top speed?
    2. What distance would a wild pike cover in this time?
    3. How long does it take an adequately designed Robot Pike to reach its top speed?
    4. What distance would an adequately designed Robot Pike cover in this time?
  6. The following diagram and paragraph compares the performance of a magnetically levitated train (the Transrapid) to a conventional high speed passenger train (the ICE).
    The Transrapid is not only fast, but it can also accelerate quickly to high speeds. 300 km/h (185 mph) can be reached after a distance of only 5 km (3 miles). Modern high-speed trains require more than 28 km (18 miles) and at least four times as long to reach the same speed.
    How does the acceleration of the Transrapid compare to the acceleration of the ICE? (Note: Given the data in the text and diagram, there are three slightly different answers to this question.)
  7. During a thunderstorm, a tree is blown over into a narrow road. A truck traveling at 16.0 m/s is 32.1 m away from the tree when the driver hits the brakes. After 4.0 s of braking the front bumper of the truck hits the tree. How serious was the collision with the fallen tree? Justify your answer with an appropriate calculation.
  8. Two cars are adjacent to each other on a four lane highway. The first car accelerates uniformly from rest at 3.0 m/s2 the moment the light changes to green. The second car approaches the intersection already moving at 18 m/s and is beside the first car at the instant the light changes. It then continues driving with a constant velocity.
    1. When are the two cars side by side again?
    2. When do they have the same velocity?
  9. In soccer, the ball is placed 11 m from the goal line during a penalty kick. A good player can kick the ball at 29 m/s. Determine the acceleration required of the goalkeeper to block this ball given that the width of the goal is 7.32 m (8 yards).
  10. A custom 1972 Vauxhall Victor modified by Andy Frost of Wolverhampton, UK is reported by some to be the world's fastest street legal car. That's not quite right. It doesn't have the highest top speed, but it probably does have the highest acceleration. Read this transcript of a report from the motoring magazine show Fifth Gear.
    Meet Andy Frost, a 45 year old automatic transmission specialist and creator of Red Victor 1. [Rapid cuts between shots of the car.] This has a 9.3 litre V-8. It's got 2,200 brake horsepower and does nought to sixty in one second. That's one second. Still not impressed? Watch this. [Telephoto shot of a quarter mile test.] The McLaren F1 can do the quarter mile in 11.1 seconds. This does it in 7.8 -- a record.
    Red Victor 1 Quarter Mile Time: 7.80 s
    Note:
    • John Lingenfelter reached a top speed of 254.76 mph (410.00 kph, 113.89 m/s) in 1989 driving a Callaway Corvette Sledgehammer. This is probably the fastest street legal car that ever existed.
    • The McLaren F1 is often used as the standard for high performance cars, but the Bugatti Veyron 16.4 outperforms the McLaren on most tests. The Bugatti has a reported nought to sixty time of 2.5 seconds. Red Victor 1 beats this easily, so it is probably the car with the greatest acceleration.
    Given all of this information, determine the following quantities for Red Victor 1.
    1. Determine the average speed during the quarter mile run in …
      1. mph
      2. km/h
      3. m/s
    2. Determine the final speed during the quarter mile run (assuming uniform acceleration) in …
      1. mph
      2. km/h
      3. m/s
    3. How do these numbers compare to the top speed of the Callaway Corvette Sledgehammer?
    		 
    1. Determine the average acceleration during the nought to sixty test in …
      1. m/s
      2. g's
    2. Determine the average acceleration during the quarter mile run in …
      1. m/s
      2. g's
    3. Why do these two calculations yield noticeably different values and what does this say about the results of your final speed calculation in part b?

calculus

  1. An object's position is described by the following polynomial for 10 s.
     
    s = t3 − 15t2 + 54t
     
    Where s is in meters and t is in seconds. Determine …
    1. the object's velocity as a function of time
    2. the object's acceleration as a function of time
    3. the object's maximum velocity
    4. the object's minimum velocity
    5. the time when the object was moving backward
    6. the time(s) when the object was at the origin (s = 0 m)
    7. the time(s) when the object returned to its starting position
    8. the object's average velocity
    9. the object's average speed
  2. An object's velocity, v, in meters per second is described by the following function of time, t, in seconds for a substantial length of time …
     
    v = 4t (4 − t) + 8
     
    Assuming the object is located at the origin (s = 0 m) when t = 0 s determine …
    1. the object's position, s, as a function of time
    2. the object's acceleration, a, as a function of time
    3. the object's maximum velocity
    4. if and when when the object stops
    5. if and when the object returns to the origin (s = 0 m)
  3. The following equations state displacement as a function of time. Derive the subsequent equations for velocity and acceleration as functions of time. (The symbols A, f, j, k, x0, π and τ are all constants.)
    1. x = ⅙jt3
    2. x = A sin(2πft)
    3. x = x0e−t/τ
  4. A crude mathematical model of tunneling is represented by the equation …
       
    v =  k
    x
       
    where v is the tunneling speed; x is the length of the tunnel; and k is a constant.
    1. In what way (increase or decrease) does the tunneling speed change as the tunnel gets longer? What engineering aspect of tunneling is causing this change?
    2. Determine the following quantities as a function of time …
      1. tunnel length
      2. tunneling speed
      3. tunneling acceleration
      (This is a somewhat difficult problem for students who have just started learning calculus.)
  5. A simplified model of a car accelerating from rest along a straight path is given by the following equation …
     
    v(t) = A(1 − e−bt)
     
    Where v(t) is the instantaneous speed of the car in feet per second, t is the time in seconds, and A and b are constants.
    1. speed
      1. What are the units in the coefficients A and b?
      2. What is the physical meaning of the coefficent A?
      3. What is the speed of the car at t = 0 s?
      4. What is the asymptote of this function as t → ∞?
      5. Sketch a graph of speed vs. time. Include the value of v(0) and the asymptote of v as t → ∞.
    2. position
      1. Derive an equation s(t) for the instantaneous position of the car as a function of time. (Be sure that your function has the value s = 0 when t = 0.)
      2. What is the asymptote of this function as t → ∞?
      3. What is the physical meaning of the slope of this asymptote?
      4. Sketch a graph of position vs. time. Include the value of s(0) and the asymptote of s as t → ∞.
    3. acceleration
      1. Derive an equation a(t) for the instantaneous acceleration of the car as a function of time.
      2. What is the acceleration of the car at t = 0 s?
      3. What is the asymptote of this function as t → ∞?
      4. Sketch a graph of acceleration vs. time. Include the value of a(0) and the asymptote of a as t → ∞.
    4. numbers
      Apply this model to a real but exceptional car -- the Red Victor 1. This car has a zero-to-sixty time of about one second and a quarter mile time of about eight seconds. In other words, let …
           
      v(1 sec) = 88 ft/sec and s(8 sec) = 1320 ft
           
      then determine …
      1. the values of the coefficients A and b [I think this can only be done using a fancy calculator.]
      2. the maximum speed, and
      3. the maximum acceleration.

statistical

  1. braking-distance.txt
    The accompanying text file contains the stopping distances in feet for 123 different automobiles. The initial speed was 60 mph in the first column and 80 mph in the second column.
    1. Determine the equation of the line of best fit for this data set.
    2. Why does the slope of this line have the value that it does?
    3. Which proportionality in this section is relevant and what value does it predict for the slope?

investigative

  1. Obtain the performance specifications for an airplane of any sort. Identify the relevant data and then calculate the tightest runway acceleration that this airplane could have and yet still take off safely. You might also need runway data for an airport capable of handling such a plane. Since there are several factors affecting take off, you should perform this calculation a few times for a range of different conditions. As an option, you could also try repeating this problem for landing instead of takeoff. That is, calculate the tightest runway deceleration that this airplane could have and yet still land safely.

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