Displacement and Total Distance Traveled

Transcription

1 Displacement and Total Distance Traveled We have gone over these concepts before. Displacement: This is the distance a particle has moved within a certain time - To find this you simply subtract its position at time zero from its final resting position. i.e. if something moves from position 3 to position 7, its displacement is 4. If it moves the other way, from 7 to 3, its displacement is -4. And if it moves from 3 to 7 and then back to 3, the displacement is zero. We can also find displacement by taking the integral of velocity. Logically, we know that this is true: new position of a particle = initial position + displacement. Since we know we can find the displacement of an object by finding the integral of the velocity function, we can now write: s() t = v( u) du + s Check it out. t position at t = equals initial position. s() = v( u) du + s = s t ds d And v= = vu ( ) du + s = vt () dt dt This gives us a couple of different ways to express position, in addition to our previous knowledge that said we could take the antiderivative of v (t) to get s (t), and it will come in handy when we can't actually find the antiderivative of the v (u). Don't forget that Total Distance Traveled is the integral of the absolute value of velocity. You must make sure that the positive and negative distances traveled do not cancel each other out. If you are allowed to use your calculator you can use it to evaluate the integral of v, but without it you must first determine where the velocity function is equal to zero and then check the signs of velocity in the subintervals on the velocity number line to see where it is positive (the object is moving up or to the right) and where it is negative (it is moving left or down) and then add the absolute value of the integrals of those sub-intervals. In general, integrals can be interpreted as net change over time. Any time we have a differential equation showing the rate of change, the integral of that function will give us the net change in the variable in that equation. The integral of ds/dt gives us net change in s. The integral of dv/dt gives us net change in v. The integral of dc/dt, where dc/dt is the rate of cola consumed over time, gives us the net change in the amount of cola consumed (or in other words, it gives us the total amount of cola consumed in the time period indicated by the limits of integration, since the rate of consumption must always be positive).

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3 Net Change Remember that the integral will give you the TOTAL under the curve; the total of whatever variable is shown on the vertical axis. If my equation gives dc/dt =... (the amount of coke consumed per hour) and C is the amount of coke consumed at time t, then the integral of C from t = 1 to 4 will tell me the total amount of coke consumed from t = 1 to 4 years. If I divide by the length of the interval, I get the average amount of coke consumed at time t within that interval. If my equation gives me dr/dt =... (the change in the amount of rats over time), where R is the number of rats in the sewer system, then the integral of R gives the total amount of rats from t = a to b. Divide by the length of the interval, we have the average number of rats within that time interval. With any equation, dp/dt =... (the rate at which P is changing over time), an integral gives the area under the curve and that area will indicate the total amount of P that was consumed, earned, whatever, within that time interval. If you divide by the length of the interval, you get the average amount. Here are some examples: A - The density of cars (number of cars per mile) on 1 miles of the highway approaching Disney World is given by the equation F (x) = 1 [ 4 - ln (x + 3) ], where x is the distance in miles from the entrance to Disney World. To find the total number of cars on the road in this 1-mile stretch, all I have to do is take the integral between x = and 1. The integral gives the area = total amount of F (cars). Note: although they did not write df/dtd =, they did define F (x) as the number of cars per mile, so it is a rate. B - A rumor is spreading at the rate of f (t) = 1 e -.t new people per day. To find the number of people who hear the rumor in days 4 to 6 we just have to integrate that rate, f (t), from t = 4 to t = 6. Again, the integral = area under the rate curve = total number of people who hear the rumor in that interval. Note: Again, even though the equation was f (t) = we are told that f (t) is the rate of the spread of rumor. A rate is the same as if I had written the equation df/dt = The same applies to example C.

4 C - An epidemic is spreading through a city at the rate of f (t) new people per week. If we take the integral of this rate between t = and 4, we get the total new people infected in the first 4 weeks. If we integrate between t = 1 and 3, we get the total new people infected in weeks 1 through 3. This can also be interpreted as the change in infected people between the beginning of week 1 and the end of week 4. These are common AP exam questions! They want to know if you know that the area under the curve (when the function defines a rate of change) also gives the total amount of whatever that dependent variable is within the specified interval.

5 Hooke s Law for Springs Now we look at the work done by a constant force. Work Done by a Constant Force W = Fd W = work in joules or ft-lb F = force in pounds or Newtons d = distance in feet or meters Remember that units are important when doing any of these problems. When working in SI units, work is measured in joules (a force of Newtons moved some meters). Using the British system, work is measured in ft-lb (force of pounds moved some feet). One pound is equal to approximately Newtons. A joule is one Newton-meter. You do not necessarily have to memorize these units and their relationship, but you do have to be aware of units and make sure that if you use feet, then use feet (not inches, for instance) throughout the problem. Now we will look at springs. To find the force it takes to compress or stretch a spring we look to Hooke's Law for Springs: Hooke's Law for Springs F = k x Where: F = force done in compressing or stretching the spring k = force constant for the spring x = length from spring's natural length Once the force is found for the spring, then we can find work done by integrating the force. Here are a couple of examples. Work done in Compressing a Spring b a ( ) F x dx

6 Where: F = force done in compressing or stretching the spring, found with Hooke s Law a and b = lower and upper lengths from spring's natural length that the spring is stretched A: A spring has a natural length of 1 m. A force of 4 N stretches the spring to a length of 1.8 m. Note that the x in Hooke's Law will be.8, because the spring was stretched.8 m past its normal length. a) Find the force constant: We just need to use Hooke's Law: 4 =.8k and k = 3 N/m b) Find the work to stretch the spring to m past its natural length: We have to use an integral to find this. F(x) = kx = 3x W = 3x dx = 3 x dx 1 W = 3 [ x W = 3 ( ) = 6 J We use Hooke's Law and the constant that we found in the first part to find an expression for force. Now we put the expression for force into our integral. The limits of integration are to (the lengths from stretch to a stretch of m past its normal length). I can move the constant, 3, out in front of the integral. This integral is easily solved. Remember your labels! In this problem we are using meters and Newtons, so work is labeled in joules. (I will not count it wrong if you use units of Newton-m c) Find how far a 45-N force will stretch the spring: We can use Hooke's Law to do this. 45 = 3x and then x = 1.5 m.

7 B: A force of N will stretch a rubber band cm (. m). Assuming Hooke's law applies, how far will a 4-N force stretch the rubber band? How much work does it take to stretch the rubber band this far? a) We know that a force of N will stretch the rubber band. m (notice how we need to keep units consistent). In order to find out how far a force of 4 N will stretch the rubber band, we need to use Hooke's Law to find the spring constant. =.k so k = 1 n/m. Now, if we have a force of 4 N, we know that 4 = 1 x, and the band will stretch x =.4 m. b) To find the work done to stretch the band.4 m, we have to use our integral: F(x) = kx = 1x.4.4 W = 1x dx = 1 x dx 1 W =1 [ x.4 W = 1 (.8 ) =. 8 J We use Hooke's Law and the constant that we found in the first part to find an expression for force. Now we put the expression for force into our integral. The limits of integration are to.4 (the lengths from stretch to a stretch of.4 m). I can move the constant, 1, out in front of the integral. This integral is easily solved. Remember your labels! In this problem we are using meters and Newtons, so work is labeled in joules. That is all there is to Hooke s Law; I think it is usually pretty easy to use. You must have this memorized. It does show up on the AP exam and on our test I will not give you the formula.

8 Homework Examples #1, 7 #1: The function v (t) = 5 cos t, where velocity is measured in m/sec for a particle moving along the x-axis. (a) Determine when the particle is moving to the right, to the left, and stopped, (b) Find the particle's displacement for the time interval t π, and (c) Find the total distance traveled by the particle. v (t) = 5 cos t t π a) Set v (t) = cos t = to find the CVs of t = π 3, π. The ends of our interval have to be included too: t =, π This is our function and interval. (a) The particle moves to the right when v >, moves to the left when v <, and is stopped when v =. Now draw the v number line and put its signs over each subinterval. Once you do that, you can answer the questions. π The particle moves right on < t < and 3π < t < π because v is positive on those intervals. π 3π The particle moves left on < t < because v is negative on those intervals. π 3π The particle is stopped when t =, because v = at those points. b) π Displacement = 5cos t dt [ t] Displacement = 5sin = π π / 3 π / π c) T. D. = 5cost dt + 5cost dt + 5cost dt π / 3 π / [ ] [ ] [ ] π / 3 / π / 3 π / TD.. = 5sin t + 5sin t + 5sin t (b) Displacement is the integral of velocity. Total Distance Traveled is the sum of the absolute values of the integrals for each section of velocity. The middle section must have its absolute value

9 [ ] [ ] [ ] TD= + + = taken because velocity is negative in that interval.

10 #7: The function v (t) = e sin t cos t, where velocity is measured in m/sec for a particle moving along the x-axis. (a) Determine when the particle is moving to the right, to the left, and stopped, (b) Find the particle's displacement for the time interval t π, and (c) Find the total distance traveled by the particle. v (t) = e sin t cos t t π a) To answer this question we will need to know where v is positive or negative, so let s use our tried and true method of discovering that information. Since e sin t > for all t, we only have to consider the second factor on this part. This is our function and interval. (a) The particle moves to the right when v >, moves to the left when v <, and is stopped when v =. 3 Set v (t) = cos t = to find the CVs of t = π, π. The ends of our interval have to be included too: t =, π Now draw the v number line and put its signs over each subinterval. Once you do that, you can answer the questions. π The particle moves right on < t < and 3 π < t < π because v is positive on those intervals. π 3π The particle moves left on < t < because v is negative on those intervals. The particle is stopped when at those points. π b) Displacement = e u = sin t du = cos t dt u u e du = e + C sint sint π Displacement = e = cost dt π 3π t =, because v = (b) Displacement is the integral of velocity. I had to use substitution to integrate.

11 π / 3 π / π sint sint sint c) T. D. = e cost dt + e cost dt + e cost dt π / 3 π / sint π / sint 3 π / sint π TD.. = e + e + e π / 3 π / [ ] TD.. = e 1 + e e + 1 e = e e Total Distance Traveled is the sum of the absolute values of the integrals for each section of velocity. The middle section must have its absolute value taken because velocity is negative in that interval.

12 Homework Examples #11, 13, 15 #11: Recall that the acceleration due to Earth's gravity is 3 ft/sec. From ground level, a projectile is fired straight upward with velocity 9 ft per sec. a) What is its velocity after 3 seconds? b) When does it hit the ground? c) When it hits the ground, what is the net distance it has traveled? d) When it hits the ground, what is the total distance it has traveled? a (t) = -3 initial velocity = 9 a) v( t) = a( t) dt = 3dt = 3t + C At t =, v = 9 9 = + C v (t) = -3t + 9 v (3) = = -6 ft/sec b) s() t = v() t dt ( 3t 9) = + = 16t At t =, s = = + + C dt t C st ( ) = 16t + 9t st ( ) = 16t + 9t = t( 8t 45) = when t =, 45/8 sec Hits the ground after 45/8 sec 45/8 c) Displacement = ( 3t + 9) dt 45/8 Displacement = 16t + 9t = d) Time to top = (45/8)/ = 45/ s = = ft Total distance = (16.565) = ft This is our acceleration function and initial conditions. It is negative because gravity is pulling the object down. (a) Velocity is the integral of acceleration. Use initial conditions to find the value of C. Evaluate for t = 3. The negative means that the object is going down when t = 3. (b) The object hits the ground when height =, so we have to integrate again. Use initial conditions to find C. It was fired from the ground, so its initial position was zero. Set position = and solve for t. The first value is when the particle started up, the second when it hit the ground. (c) The net distance is the same as displacement, the integral of velocity over the interval. Since the starting and ending height are the same, the net distance is zero. Total Distance Traveled is the distance to the top of the arc plus that same distance traveled down to the ground. The time for each of these intervals will be half the time we got in (b). I will find the distance to the top, and then use symmetry to get the total distance.

13 #13: We are given a graph in the book that shows the curve of velocity. The numbers in each section of the curve are the area of the region. What is the total distance traveled by the particle in the time period? Total distance traveled = sum of the absolute values of the area of each region Total distance traveled = = 33 cm #15: Using the same graph as in #13, approximately where does the particle achieve its greatest positive acceleration on the interval [, b]? Acceleration is the derivative of velocity, so it also refers to the slope of the curve of velocity. Thus we need to find the greatest positive slope within the interval [, b]. This occurs at a.

14 Homework Examples #17, 19 #17: The graph of the velocity of a particle moving on the x-axis is given in the book. It consists of four slanted line segments. The particle starts at x = when t =. (a) Find where the particle is at the end of the trip. Where the particle is at the end of the trip is displacement. In this case the function is always positive so there are no negative regions to worry about. Position within any interval can be found by taking the integral of the function within that interval or in this case, by using geometry and finding the area under the curve. There are four triangles so the area under the curve is 4 ( ½ (1)() ) = 4 Initial position is. Thus the particle is at + 4 = 6 (b) Find the total distance traveled by the particle. It started at and traveled 4 units (found in (a)) so total distance traveled is 4. #19: The graph of the velocity of a particle moving on the x-axis is given in the book. It consists of a number of joined line segments. The particle starts at x = when t =. (a) Find where the particle is at the end of the trip. Where the particle is at the end of the trip is displacement which means that positive and negative directional travel will be allowed to cancel each other out. Position within any interval can be found by taking the integral of the function within that interval or in this case, by using geometry and finding the area of the region. There are three regions to look at. From to 1 we have a triangle. It is below the axis, so its area is negative: -½ (1)() = -1 From 1 to 5 we have a trapezoid above the axis: ½ ()(1 + 4) = 5 From 5 to 7 we have a triangle below the axis: -½ ()(1) = -1 Initial position is. Thus the particle is at = 5 (b) Find the total distance traveled by the particle. This time all areas must be positive so that we can get the total distance traveled. It started at and traveled so total distance traveled is 7.

15 Homework Examples #1, 9 #1: The rate of consumption of oil in the United States during the 198's (in billions of / 5 barrels of per year) is modeled by the function C = 7.8 e t, where t is the number of years after January 1, 198. Find the total consumption of oil in the United States from January 1, 198 to January 1, 199. Total consumption will be found by the integral of the rate of consumption within that interval. We will integrate from t = to t = 1. Total oil consumption = 1 t / 5 t/ 5 1 /5 7.8 e dt = (7.8)(5) e = (7.8)(5) e billion barrels #9: A certain spring requires a force of 6 N to stretch it 3 cm beyond its natural length. a) What force would be required to stretch the spring 9 cm beyond its natural length? First we find the force constant by using Hooke's Law. We are told that it takes a force of 6 N to stretch a spring a total of 3 inches. That means that: F = kx 6 = 3k k = Now we can find the force needed to stretch the spring 9 cm. F = (9) = 18 N b) What would be the work done in stretching the spring 9 cm past its normal length? N-cm W = x dx = x =