Physics 21: Homework 1 Hints

Transcription

1 Physics 21: Homework 1 Hints 1. This is a situation in which the car is clearly not moving with a constant acceleration. (a) Here, t = 0 is presumed to be the initial time for which this motion is monitored. (b) Note that the units of αt 2 and βt 3 must be in meters. Since the units of time are in seconds, then the units of α and β immediately follow. (c) Note that x (t) undergoes some interesting features, so that the car s movement is nontrivial. i. Recall that the definition of the average velocity typically implies looking over a reasonably large time scale. In particular, v = x t = x (t f) x (t i ). t f t i Since the functional form for the position is known, the average velocity may be found by evaluating the ratio at the times being considered. For example, for A, t i = 0 and t f = 2.00 s. ii. Here, because we are seeking for instantaneous forms, we must take derivatives. In particular, recall that v = dx dt, and that a = dv dt. iii. To look at such points, known as turning points, one must investigate whether or not the velocity undergoes a sign change. Indeed, the velocity contains information not only pertaining to the speed with which the car is traveling, but also the direction of the car s movement. Since x (t) is a very smooth function, this means that the velocity will also be quite smooth. For that reason, in order for the velocity to switch sign, it will smoothly have to go through a value of zero. For this reason, determining such turning points amounts to determining the points at which the velocity is zero. Upon finding the times corresponding to such zeroes of the velocity function, one can then determine if, indeed, the car physically switches sign from before that zero to after that zero. iv. A really neat way to graph these functions is to investigate the dependence of the functions and small and large values of time, as there is going to be a competition of terms, particularly for x (t) and v (t). For example, note that for small values of time for x (t), αt 2 is going to be much larger than βt 3, so that the functional form for x (t) for values near t = 0 will be that of a parabola opening upward. For large values of t, the dominant terms will then become βt 3, but because of the negative sign in front, it will switch the functions concavity to a negative value. Thus, x (t) has a local maximum at some intermediate time, and must eventually cross x = 0 at some later time. Similar analysis may be done for v (t). The plot for a (t) is quite straightforward, as it s a linear plot given the cubic nature of x (t). 2. For this problem, it is imperative to remember that the slope of x (t) represents the velocity. In particular, the sign of this slope indicates the direction of travel (either positive or negative), while the value of this slope indicates the speed (which is a non-negative number). (a) To analyze this, one must look at points at which the slope of the x (t) plot is zero. In other words, wherever the tangent line at a point is horizontal, the speed is zero. (b) If the region between successive points in the x (t) plot undergoes an unchanging slope, then the speed is constant in that region. (c) If the region between successive points in the x (t) plot undergoes an increase in the absolute value of the slope in time, then the speed is increasing in that region. (d) If the region between successive points in the x (t) plot undergoes a decrease in the absolute value of the slope in time, then the speed is decreasing in that region. 1

2 (e) When the slope at a point on the x (t) graph is positive, then the velocity is also positive. In other words, such a positive slope indicates that the object is moving in a positive direction (e.g., East, if East is defined as positive). Even though it is possible for the speed of an object to decrease either moving positively or negatively, this particular part is interested in the case in which the speed of the object is decreasing while the object is moving positively. (f) When the slope at a point on the x (t) graph is negative, then the velocity is also negative. In other words, such a negative slope indicates that the object is moving in a negative direction (e.g., West, if East is defined as positive). Even though it is possible for the speed of an object to increase either moving positively or negatively, this particular part is interested in the case in which the speed of the object is increasing while the object is moving negatively. 3. The relevance of the black-out acceleration quoted in this problem is to describe the trauma felt by a person s head as it slows down in the process of a car-crash when the car happens to hit, say, a brick wall while moving forwards. (a) Although notions of inertia will be discussed in the near future in the class, this partly is mostly concerned about an experiential explanation. Consider the sensation one feels when suddenly braking in a car, for example, when a stop-light suddenly turns yellow and you find that it is necessary to stop the car abruptly to avoid running into the intersection onto traffic that is eager to make a left-turn from the opposing lane of traffic. Under these circumstances, you feel as though your body, as a whole, wants to keep moving forward. Of course, what restrains your body is your seat belt. Since all parts of your body want to keep moving forward, the same applies to your eyes. In this respect, the event is an eyeballs-out experience, because if your eyes were not held in place by your optic nerve, eye muscles, eye lids, and skull sockets, then your eyeballs would theoretically fly out if your head proceeds to stop moving. (b) The process in this part involves the car to slow down from 100 km/h to 0 in a fairly abrupt fashion. Although the assumption of the deceleration being constant is not entirely true, it nonetheless paints a picture of the relevant time-scales involved in such a process. i. Applying kinematics at a constant acceleration, the time taken is 43.2 ms. ii. Note that in order for the airbag to function properly, it must pretty much inflate within the time found in Part (bi), because it will likely be not much after this time that the person s head comes into contact with the inflated airbag. The person s head will roughly be moving at 100 km/h upon impacting the inflated airbag. Thus, the airbag s purpose is to slow the person s head down over a distance that is ideally as long as possible in order to keep the person from blacking out. Nonetheless, the minimum distance over which this head must stop is that which provides the head with an acceleration that is at the black-out value. The answer is: 1.93 m. 4. In this problem, note that the Veyron does not accelerate uniformly, for if it did, then the time, for example, to reach 300 km/h from rest would be three times the time it would take to reach 100 km/h from rest. (a) In this part, use the fact that i. Here, v = 100 km/h and t = 2.46 s. ii. Here, v = 200 km/h and t = s. a = v t. (b) This part presumes that the values of the average acceleration found in Parts (ai) and (aii) are going to serve as the acceleration at all times for the time intervals of t = s and t = 2.46 s 14.6 s, respectively. Thus, the kinematic equations for constant acceleration are to be used for the analysis in the sub-parts. 5. Notice that A starts ahead of B initially, that A starts from rest and accelerates, and that B is already moving at t = 0 and never changes its speed throughout the process. (a) The positions of the trains are identical wherever their x (t) graphs intersect. 2

3 (b) This part requires some reasonable estimation of velocities from the position graphs of the trains. The velocity graph of B simply requires an estimation of the slope of the linear plot of x B (t). As for A, note that the slope of x A (t) is practically zero at t = 0. Moreover, note that, by the Mean Value Theorem of calculus, that there is a point between the points of intersection between A and B at which the slope of A is the same as the slope of B. Since it s presumed that x A (t) is parabolic, then its velocity must be linear. As such, by utilizing these two data points of the slope of A s position graph, one can then determine v A (t). (c) The discussion of such a notion was described in Part (b) above. (d) Whenever the position graph of A becomes larger in value than that of B is when A passes B. (e) Whenever the position graph of B becomes larger in value than that of A is when B passes A. 6. The main thing to take away from this problem is that the area under the curve of the v (t) plot determines the position change referenced from the start of the interval being considered. In this plot, when v > 0, the object is moving in the positive direction; whereas, when v < 0, then the object is moving in the negative direction. For this reason, the slope of the x (t) plot must be consistently positive up to the 5-s mark. Beyond that, the slope should be negative. Of course, the slope should be zero right at t = 5 s. Moreover, the plot for x (t) must not exhibit any kinks (i.e., cusps), because v (t) is continuous; however, because v (t) is only piecewise smooth, the acceleration will undergo discontinuities at the kinks: namely, at t = 5 s and t = 8 s. Finally, because no information is provided as to the position of the particle at t = 0, then there is no way to know the absolute position of the particle. For this reason, x (t), is undetermined up to some integration constant. 7. To obtain all relevant kinematic plots, the graph must be integrated once to obtain v, and another time to obtain x. Because we are told that v (t = 0) = 0 (since the particle starts from rest) and that x (t = 0) = 20 m (since West is taken as positive and the particle starts off 20 m East of the origin), then we know the exact values of the velocity and position at any time, t, from the information provided from a (t). Although it is straightforward to determine the area under the curve of the a (t) plot (since it simply involves areas of triangles), the area under the v (t) curve will be less straightforward, because, for example, for t [0, 4 s) and for t [4 s, 6 s), the curves will be parabolic. Thus, to get exact graphs for x (t), it is necessary to properly represent the piecewise functional forms of a (t), and then use the piecewise functions to integrate twice to obtain the exact form for x (t). Although the process is somewhat involved, just remember how to express the equation of a line, because this will be the starting point to determine the functional form of a (t) for the intervals t [0, 4 s) and t [4 s, 6 s), which will be used to integrate to find v (t) for those intervals, and then, finally, for x (t) in those intervals. 8. This problem has the same inner-workings of Problem 7, except that it is considerably simpler since the functional form of the acceleration is straightforward and already provided. Note that v (t = 0) = m/s since the object is traveling positively at 20.0 m/s. (a) Since the units of the acceleration are in m/s 2, and since the time is measured in units of s, then this puts a condition on the units of γ. (b) The information states that v (t = 15.0 s) = 0. i. From the form of the acceleration, note that the acceleration opposes the velocity of the object initially. Based on what has been covered in class, it should be clear why the object reaches an instantaneous point of rest eventually. ii. Upon integrating a (t), and using the fact that v (t = 0) = 20.0 m/s and that v (t = 15.0 s) = 0, then knowledge of the value of γ immediately follows. (c) This is straightforwardly determined by integrating v (t). (d) Because it was reasoned in Part (bi) that the object reverses direction at t = 20.0 s, then since 20.0 s > 15.0 s, then calculating distance will not amount to taking the absolute value of the position change from t = 0 to t = 20.0 s. Instead, the position change must be calculated for each leg of the movement (one for the movement in the positive direction and the other for the movement in the negative direction), and then the absolute values of the position changes for those legs must be added together. 3

4 9. The key to this classic problem is to be very careful about what the meaning of traveling at two different speeds in a trip is when the traveling speeds are described as happening in terms of length versus in terms of time. (a) In this case, because the speeds are shifted abruptly at the halfway point in length, then it should be clear that the time taken to travel at the lower speed must be longer than the time taken to travel at the higher speed. In other words, there is an imbalance in time. Of course, the time for each leg can be added together to find the total time of travel. Since the average velocity is defined as the ratio of the position change to the time over which the change occurred, then the average velocity for the entire trip must consider the total position change (12 km) and the total travel time. However, because the length of each leg is known, then the time of each leg can be determined, allowing one to determine the total travel time. Intuitively, it should be expected that the average velocity is less than the average of the two velocities, because the lower-speed leg took a longer time than the higher-speed leg, so that the average velocity should be skewed towards the slower velocity. (b) In this case, because the speeds are shifted abruptly at the halfway point in time, then, by construction, the average velocity should just be the average of the two velocities. To prove this, it is worthwhile to consider that since 30 km/h > 10 km/h, then when the bicycle is traveling slower, it will cover less distance than when it is traveling faster if the travel times are the same for each of the legs (which, of course, they are). Thus, the transition from the slower to the faster speed must occur at a length substantially smaller than the midpoint of the journey. Of course, there are two unknowns here: the point in space where the transition in speeds occurs, as well as the total travel time. However, those two unknowns can be determined using two equations, the easiest of which are the uniform-motion equations for each leg. These results can then be used to confirm that the average velocity is the average of the two velocities. 10. This is a straightforward application of kinematics at a constant acceleration under the conditions of free fall. (a) The answer is: 5.60 m/s. (b) The answer is: s. (c) The answers are: 8.28 m/s and s. 11. This is, again, a straightforward application of kinematics at a constant acceleration under the conditions of free fall. (a) The answer is: 30.6 m. (b) The answer is: 24.5 m/s. (c) These graphs are dependent upon the chosen coordinate system (i.e., whether up is taken as positive or negative ). Nonetheless, the acceleration graph is trivial. The velocity graph should be linear and should be zero at t = 0 (since the rock is dropped). Finally, the position graph should be quadratic and have a monotonically increasing slope magnitude (starting with zero slope). 12. The key idea to this problem is that since all parts of the hot-air balloon are moving at a common velocity specifically, in an upward direction at a speed of 10.0 m/s then so long as the sandbag is released gingerly (i.e., with no additional downward push or pull), then the sandbag must have an initial velocity of 10.0 m/s upward upon disconnecting from the balloon. Thus, this problem boils down, for example, to one where a rock is thrown upward at some height above the ground. (a) The answer is: 55.1 m. (b) The answers are: 53.8 m and 50.4 m. (c) The answer is: 4.37 s. (d) The answer is: 32.9 m/s. (e) These graphs are dependent upon the chosen coordinate system (i.e., whether up is taken as positive or negative ). Nonetheless, the acceleration graph is trivial. The velocity graph should be linear and should be ±10 m/s at t = 0 (since the rock is thrown vertically). Finally, the position graph should be quadratic and have a monotonically decreasing slope magnitude (starting with zero slope) up to the time at which the maximum height occurs, after which it should have a monotonically increasing slope magnitude. 4

5 13. The key for this problem is to note that it is not initially given what the details of the brick s movement is after being dropped. Even though one is typically tempted to want to uncover those details from the get-go (since we typically like to know all of the pieces of information initially), it is actually not necessary whatsoever. Certainly, the brick is traveling at some speed, call it v 0, at the beginning of the 2.0-s over which it travels 80.0 m. This initial speed may be determined using y (t) = y 0 + v 0 t at2. In turn, one may then find the final speed at the end of this 2.0-s interval using any of the remaining constantacceleration kinematic equations. Finally, this latter final speed serves as the initial speed for the upcoming 2.0-s interval. Thus, with the use of the above kinematic equation again, one will be able to determine the position change of the brick in that subsequent time interval. 14. Even though across the whole journey of this train the acceleration is not constant, it is constant for each piece of the journey. Thus, the problem must be treated with the constant-acceleration kinematic equations for each respective regime in a piecewise fashion. For example, the speed of the train at t = 20.0 s is the same up to t = 100 s, so that the final speed of the initial speeding-up regime is the same as the initial speed of the final slowing-down regime. The train never changes direction, so the position change always serves as the traveled distance in all regimes. The answer is: km. 15. As discussed in class, this is a typical problem for which one wants to find the intersection of two position graphs: one for the man which is linear graph with, say, positive slope, and another for a bus which is a quadratic function with zero initial slope that has a head start and maintains a monotonically increasing, positive slope. For arbitrary parameters, the collision time interpreted as the times at which the graphs intersect, if at all could have two real solutions, one real solution, or two complex-conjugate solutions (i.e., no solutions). The nature of the result is determined by the discriminant of the quadratic formula, which is the stuff under the square-root of that formula. It will be found that no solutions exist for Part (a), but that two real solutions exist for Part (c). Thus, since no solutions exist for Part (a), one needs to find the closest the man got to the bus, which amounts to comparing the position functions of each entity and finding the point in time when the difference between the functions is the smallest in value. Moreover, note also that the minimum speed required for the man to catch the bus actually coincides with the case for which the point of intersection is at a single point (i.e., that only one real solution exists from the collision requirement). 16. This problem is truly an alternate version of Problem 15. One may treat this problem by imagining that the front of Train A is, say, Particle A, while the back of Train B is Particle B. The problem is to see if the two particles will collide (i.e., be at the same location at the same instant of time) given the provided circumstances. Note that if both particles are taken to be traveling, for example, in the positive sense, then the acceleration of A is in the negative sense. Whether or not the particles collide will, again, be determined by the discriminant of the quadratic formula that arises from imposing the collision condition, namely x A (T ) = x B (T ), where T is the collision time. Part (d) of the problem is a play on the alterations done in class regarding the problem on a cop catching a speeder. The part is quite open-ended, but a good way to approach the possible renditions is to look at how to alter the graphs of the problems to get unique intersecting features. Of course, one can always add a narrative to the plots to make sense of the circumstances for which the position plots may be reproduced in a real-world example. 17. We always think of a ramp as something that is neither horizontal nor vertical. As such, we are tempted to believe that when a ball rolls on a ramp that it is moving in two dimensions. Indeed, we believe that since the ball rolls down a ramp, it must be moving horizontally, say, to the right and moving vertically down. That being said, a ramp has a flat surface. If one suspends one s disbelief for a moment by viewing a ramp by tilting one s head in such a way that the ramp looks horizontal to one s perspective, then a ball rolling down such a ramp looks to be moving horizontally in such a view. Thus, it makes no sense to try to convince oneself that a ball rolling down a ramp is two dimensional. Indeed, since the ball rolls along a straight line, it is inherently one dimensional. For this reason, this problem may simply be treated as a one-dimensional problem. In particular, 5

6 we assume that the motion of this marble is with a constant acceleration, with the movement of the ball purely constrained to the line defined by the ramp itself. As such, the constant-acceleration kinematic equations may be used, with, for example, the x-axis taken as an axis that lies along the straight-line defined by the ramp. Assume that +x corresponds to moving down the ramp, so that both the ball s velocity and its acceleration are directed along +x. Now, because the value of the ball s constant acceleration is unknown, note that this is as good a point as any to use the equation, x x 0 t = 1 2 (v + v 0), since this equation is primed to be used in such an instance. Although there are a few ways to solve this problem, a nice way to approach it is to use the fact that since the object starts from rest, then after the second 5.0-s interval, the speed of the ball will be twice the speed of the ball at the conclusion of the first 5.0-s interval. Thus, by using the above equation, for example, for the distinct intervals of t = s and t = 5.0 s 10 s, then the answer will follow. (a) The (exact) answer is: m. (b) The (exact) answer is: 8 3 m/s Both objects must be represented with the use of a common coordinate system. The choice of the coordinates is totally arbitrary (i.e., there is freedom to choose the position of the origin and the meaning of positive); however, the choice must be cemented and used without wavering from it so that inconsistencies do not arise. As long as such a prescription is followed, the final answer will turn out to be the same independent of the choice of coordinates. (a) The answer is: h/v 0. (b) One must be careful in this part, because the point of the problem is to have the object s collide prior to the point in time at which the thrown ball reaches the ground. It helps to consider the ground as some imaginary horizontal line. Before adjusting the height of the dropped ball, it helps to see how the adjustment of the velocity of the thrown ball affects the result when the height is fixed. In particular, note that if the thrown ball is shot with an enormous speed upward, the collision will occur practically at the height at which the ball above it is dropped. If that speed is somewhat reduced, then they will likely collide at a lower height above the ground. However, note that if the upwardly thrown speed is very small, the ball on the bottom will practically reach the ground instantaneously, so that there would be no collision above the ground. (For that matter, there actually wouldn t be any collision at any point even below the ground, if the ground was envisioned as some imaginary horizontal line.) A similar game may be played by keeping the thrown speed constant but adjusting the height, as is the case in this part. If the height of the dropped ball is practically right above the thrown one, then they will clearly meet while above the ground. If the height is huge, then the thrown ball will reach its maximum height likely without colliding with the dropped ball. At that point, although the dropped ball will have a larger speed than the thrown one (since the thrown one has zero speed at its maximum height), the height differential may still be too large for the dropped ball to catch up to the thrown one before reaching the ground. Thus, there is a compromise, and the largest height from which the dropped ball may be released would result in the two colliding right at the moment that both reach the ground. Imposing such a condition, one finds the answer to be: 2v 2 0/g. 19. It is presumed that the AC unit fell from rest, so that this piece of information must also be integrated into the solution as well. For the regime during which the AC unit falls across this window, one may determine the speed of the unit upon reaching the top of the window. In turn, this speed may be used as the final speed for the segment when the AC is seen to break off. As such, the answer becomes: m. 20. The key to this problem is to realize that there are two relevant accelerations: one for which the rocket is speeding up from launch all the way up to the point where it loses its fuel supply, and another for which the rocket is in free fall. Although the upward acceleration of the rocket (which included the effect of gravity) up to the point where it depletes its fuel is unknown in value, it is assumed to be constant. Thus, this is a quintessential problem for which technically the acceleration is not constant through the entire flight of the 6

7 rocket, but is, rather, piecewise constant. As such, two sets of constant-acceleration kinematic equations must be utilized for each of the regimes. The two sets are connected by a couple of parameters. Firstly, the final position at which the fuel is depleted serves as the initial position for which the rocket begins its free fall; secondly, the final velocity of the rocket at which it loses its fuel supply serves as the initial velocity of when it goes into free fall. We know the duration of the upward-acceleration regime, and we know a data point of the height of the rocket (relative to the ground) during a specific point in the free-fall regime (although that specific data point is not the maximum height of the rocket, as one could quite easily show from the result of Part (b)). Ultimately, there are three crucial unknowns the upward launching acceleration, the final velocity of the upward-acceleration regime (or, equivalently, the initial velocity of the free-fall regime), and the final altitude of the upward-acceleration regime (or, equivalently, the initial altitude of the free-fall regime) in which case it should be clear that the constant-acceleration kinematic equations in one acceleration regime are insufficient to solve the problem. Indeed, recall that for one acceleration regime, there are two independent equations which, thus, cannot uniquely determine three unknowns. (a) The answer is: 22.1 m/s 2. (b) The answer is: 393 m/s. (c) The answers are: 14.4 km & 65.1 s. (d) The answer is: 119 s. 7