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Unit1 Exmples 2.docx - l=t$t=tfirno ll Ill.l nit 3|...

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Unformatted text preview: l=t$t=tfirno ll Ill.l nit 3|] all .‘1ll El 11 ‘t (9-) Tangent. at P1 whose x (m) slope is 112: 21.0 nits FIGURE 1-]! Example 2-3. (a) Engine traveling on a straight track. (b) Graph otxvst: J: = Alt2 + B. Given 1: as a function of t. A jet engine moves along an experimental track (which we call the J: aids) as shown in Fig, 2—13a. We will treat the engine as if it were a particle. Its position as a function of time is given by the equation I = Al2 + B, where A = 2.1tlm1’s2 and H = 2.80m, and this equa- tion is plotted in Fig 2—13b. (1:) Determine the displacement of the engine during the time interval from t1=3.005 to I»; = 5.00s. {b} Determine the average velocity during this time interval. (17) Determine the magnitude of the instanta- neous velocityI at t = 5.00s. APPROACH We substitute values [or t1 and t2 in the given equation for x to obtain x1 and 1:3. The average velocity can be found front Eq. 2—2. We take the deriva- tive of the given 1: equation 1with respect to t to find the instantaneous velocity, using the formulas just given. SOLUTION [a] At :1 = 3.00s, the position (point P1 in Fig.2—13b] is 11. = .41? + B = (atom/1mm? + 2.30111 = 21.7111. At t; = 5.00 s, the position (P1 in Fig. 2—13h) is x1 = (2.11::111/11=}{s.tio1)l + 2.80m = 55.3111. The displacement is thus I; — :1 = 55.3rn — 21.7m = 33.6m. {h} The magnitude of the average velocity can then be calculated as 11.11 x1 - x1 33.6m ”=E=l2-t1 Ems This equals the slope of the straight line joining points P1 and P2 shown in Fig. 2—13b. (e) The instantaneous velocity at t = t1 = 5.003 equals the slope of the tangent to the curve at point P: shown in Fig 2—13b.We could measure this slope off the graph to obtain 1);. But we can calculate an more precisely for any time i. using the given formula x=A12+B, = 1-5.8 m/s. which is the engine’s position x as a function of time t. We take the derivative of x with respect to time (see formulas at bottom of previous page): _£=£ 1 = 11—!” d!(ar+a) am. We are given A = 2.10nt/s3, so for .t = t2 = 5.00s. 113 = 2.41 = 2(2.10mfs2](5.00s.) = atom/1. Acceleration given I“). A particle is moving in a straight line so that its position is given by the relation J: = (2.1K! mfszit2 + (1mm), as in Example 2—3. Calculate (it) its average acceleration dun'ng the time interval from t, = 3.0th to II = 5.00 s, and {it} its instantaneous acceleration as afunction of time. APPENCH To determine acceleration, we first must find the velocity at 1‘1 and 33; by differentiating x: t: = drfdf. Then we use Eq. 2—5 to find the average acceleration, and Eq. E—IS tn find the instantaneous acceleration. SOLUTION {a} The velocity at any time t is if): d u = E = E [massive]:1 + ass :11] = (4.21:: Infszjt, as we saw in Example 2—3-12. Therefore, at t1 = Sflfl s, 1:1 = {4.20 misijfiifltl s} = 12.6 mfs and at I; = 5H} s, u; = 21.!) me. Therefore, 2 21.0 nu's — 12.6 refs — = = — = 2 a as sass — sins 4‘10"”5' [b] With it = (42$! misfit, the instantaneous acceleration at any time is _ fl = i 2 = 2 a — m {it [(4.3] mf's )t] 4.2L! n1fs . The acceleration in this case is constant; it does not depend on time. figure 2—13 shows graphs of (a) .I vs. I [the same as Fig. 2—13h}, {In} 11 vs. II, which is linearly increasing as calculated above, and (c) e vs. I, which is a horizontal straight line because a = constant. Rurlwav design. You are designing an airport for small planes. Cine kind of airplane that might use this airfield must reach a speed before takeoff of at least 218 mg’s {IUD Irmfh}, and can accelerate at 2.02] mfsz. (a) If the runway is 150 m long, can this airplane reach the required speed [or takeofi‘? {b} If not, what minimum length must the runway have?I fiPPHflACH The plane’s acceleration is constant, so we can use the kinematic equations for constant acceleration. In [a], we want to find a, and we are given: Km Walled an = t] u a. = n x =15tlm e = lilting-ts1 SDLUTIDH {a} Of the above four equations, Eq. 2—12c will give us 1: when we know 1.1.3., o, x, and x0: 1:: = 1:3 + Eafix — xfi} = u + statiomissflisomj = enemies in = "tittitlv‘lniglfs2 = 24.5my’s. This runway,r length is not sufficient. [b] How we want to find the minimum length of runway, .14: — xfl, given a: = 213 mills and a = 2.01} mfsz. So we again use Eq. 2—12c, hut rewritten as { )_ nz—trfi _ (2?.flmfsf—[i " 1” 2a static naval] A EDD-m runway is more appropriate for this plane. MUTE We did this Example as if the plane were a particle, so we round off our answer to 20¢] n]. = 193 to. Falling from I tuner. Suppose that a hall is dropped [to = ll) from a tower Tiltlm high. How far will it have fallen after a time t1 = Littler; = hilt] si and r, = 3.00 s? Ignore air resistance. AFFIDACH Let us take it as positive downward, so the acceleration is a = g = +9.8llmfsz. We set on = ii and ya = (I. We want to find the position v of the hall after three different time intervals Equation 2—12h, with .r replaced by y, relates the given quantifies (t, a, and an) to the unknown y. SDLUHUN We set t = t1 = 11le in Eq.2—12h: 321 = anti + set? = [l + set? = %{9.8flmj51][l.flfls)2 = 4.90111. rIhe hall has fallen a distance of 4.90m during the time interval t=ll to II = 1.00 s. Similarly. alter 2.00s (= t1}. the halls position is gs = as = %[9.3[lmiszl(2.flfls}2 = 19am. li‘inallv1 after 3.00s [= t3). the hall’s position is (see Fig. 2—29) h = so; = emancipation? = 44.1111. Thrown down from a trim Suppose the hall in Example 2-14 is thrown downward with an initial velocit].r of 3.01} mfs, instead of being dropped. (a) What then would be its position after Mills and son s? (it) What would its speed he alter tiltls and 2.011s? Compare with the speeds of a dropped hall. APPROACH Again we use Eq. 2—12hI but now on is not zeroI it is on = Sill] mfs. SOUND“ (a) At t = 1.005, the position of the ball as given by Eq. 2—le is y = out + to? = (3.Dflm/s}[1.flils} +%[9.8[Im,’s1}{1.fllls}3 = 1.90m. At t = son a, {time interval t = I] to t = 2.00 s],the position is y = not +%ot1 = (soc monsoon +§[s.som{51}(2nos}= = 25.6w. As expected, the hall falls farther each second than if it were dropped with on = ll. (it) The velocity is obtained from Eq. 2—12a7 o = on + at = Iiilllnifs + (Eddmfsglflmsj = 12.3mfs [at a = lilils] = son this + (9.8flmfszl{2.tlfls) = memo. [sirI = sons] in Example 2—14, when the hall was dropped (on = D], the first term (on) in these equations was acre, so a = [l + at = [9.80 misllllfllls) = 9.30 mfs [at t1 = 1.005] = [v.somrstjaoos) = 19.5mfs. [at a = one s] HDTE For both Examples 2—14 and 2—15, the speed increases linearly in time by 9.3!} mfs during each second. But the speed of the downwardlyr thrown hall at any instant is always 3.00 nifa (its initial speed} higher than that of a dropped ball. FIGURE 2-19 Example 2-14. {a} An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. [See also Fig. 2—25.] (h) Graph of 3; vs r. Acceleration dueto gravity . __ ____y=fl yl=4£lflm (storms i " fl. v2=l9om H [II [After 2.005] y3=44.l m (Afteriiflilsj m Ball tltl'own ward. I. A person throws a hall upward into the air with an initial velocity of 15.0 mfs. Calculate {a} how high it goes. and {it} how long the hall is in the air before it comes back to the hand. Ignore air resistance. APPROACH We are not concerned here with the throwing action, but only,r with the motion of the hall efler it leaves the thrower’s hand (Fig. 2—31)} and until it comes back to the hand again. Let us choose 1' to be positive in the upward direc— tion and negative in the downward direction. [This is a difierent convention from that used in Example; 2—14 and 2—15, and so illustrates our options.) The acceleration due to gravity is doomward and so will have a negative sign, a = —g = —9.80 ms“. As the hall rises, its speed decreases until it reaches the highest point (B in Fig. 2—30]. where its speed is zero for an instant; then it descends, with increasing speed. SULUTIUN [a] We consider the time interval from when the ball leaves the thrower’s hand until the hall reaches the highest point. To determine the maximum height, we calculate the position of the ball when its velocity.r equals zero to = (It at the highest point}. At r = I] {point A in Fig. 2—3:!) we have Jig. = I]. on = 15.1] mfs, and a = -5|.Eiirn{'sz. At time t {maximum height], 1.! = D, a = -9.flfl mfsf, and we 1wish to find 3;. We use Eq. 2—12c, replacing it with y: a2 = a3 + Zay. We solve this equation for y: i — at = a — {15o mtg)? 2a 2|[-9.B[|mj's1} The hall reaches a height of 11.5 In above the hand. {it} Now we need to choose a different time interval to calculate how long the hall is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time reguired for the hall to reach its highest point, and then determining the time it takes to fall back down. However. it is simpler to consider the time interval for the entire motion from A to B to C (Fig. 2—3D} in one step and use Eq. 2—12h. We can do this because y represents position or displacement, and not the total distance traveled. Thus. at both points A and CL y = ft. We use Eq. 2—12h with a = —9.Bll nifs2 and find 11 = 11.5111. y = r = at+wt+iatz ll = ll + (15.0 mfalt + fl—Elfiilmfszltz. This equation is readily factored (we factor out one I}: {1511me — memoir): = c. There are two solutions: 15.1] Infs t=D dt=—=3.tlE-. an 4.9[1 this2 5 The first solution {I = [t]: corresponds to the initial point (A) in Fig. 2-30, when the hall was first thrown tom 3; = [1. The second solution, t = ans 5. corresponds to point C, when the ball has returned to y = [1. Thus the hall is hi the air 3.065. NUT E We have ignored air resistance, which could he significant, so our result is only.r an approximation to a real, practical situation. Ball thrown upward at edge of cliff. Suppose that the person of Examples 2-16, 2-18, and 2-19 is standing on the edge of a elifl, so that the hall can fall In the base of the cliff Sflflm below as in Fig. 2—32. [st] How long does it take the hall to reach the base of the clifi‘? [b] What is the total distance traveled by the ball? Ignore air resistance (likely to he significant, so our result is an approximation). APPROACH We again use Eq. 2—12h, but this time we set 3;: = —5fl.t] in, the bottom or the curt, which is. sooth below the initial position (ya = o]. SOLUTION (ti) We use Eq. 2—12h with it = -9.Bflnt,lsz, tin = liflmfs, p0 = i}, and y = -5Et.lim: y = )h + tint + that: -5t].[itn = t] + (15.0mm)! - H9130 mfszitl. Rewriting in the standard form we have (esomxsihi — {lSflmith - {seem = it. Using the quadratic formula. we find as solutions t = 5.0“! s and t = -2.01 s. The first solution, t =51)”, is the answer we are seeking: the time it takes the hall to rise to its highest point and then fall to the base of the cliff. To rise and fall back to the top of the cliff took 3.66s (Example 2—115]; so it took an additional 2.01 s to tail to the base. But what is the meaning of the other solution, t= —2.IZ11 s? This is a time before the throw. when our calculation begins, so it isn't relevant herefr (it) From Example 246, the hall moves up 11.5 to, falls 11.5 in back down to the top of the cliff, and then down another Sillirn to the base of the cliff, for a total distance traveled oi T31] rn. Note that the displacement, however. was 40.0111. Figure 2—33 shows the 1! its. 1‘ graph for this situation. Integrating a time-varying acceleration. An experimental vehicle starts from rest [on = [2-] at t = fl and accelerates at a rate given by rt = (1m mfssh. 1What is {it} its velocity and [it] its displacement 2.00s later? APPROACH We cannot use Eqs. 2—] 2 because a is net constant. We integrate the acceleration a = [111de over time to find 1: as a function of time; and then integrate t! = :11de to get the displaeement. SOLUTIUIN From the definition of acceleration, a = deficit, we have tit: = a tit. We take the integral of both sides from t: = [l at t = D to ‘treloeitgtr it at an arbi— tram»r time t: til .I' Jae Jen‘t El El 1: = thi't'llfl mfflt cit n [Tilt] mf§}(%2) I t} 2 {am whim; — o) = (soomfihz. At t = 2.00 s. t: = [3.50 mfs3}(2.t}t] s? = 14.0 nuts. {in} To get the displacement, we assume x“ = t] and start with a = deficit which we rewrite as air = it sit. Then we integrate from a: = It] at t = t] to position xattimet: 1! l de = Jodi e e 1'3 Ellis 1m: x = Jfififlmfsfltzdt = [3.Sflmfs3JE = 9.33m. U El In sum, at r = 211th, ti = 14.0m/s and x = 9.33 in. three short trips. An airplane trip involves three legs, with two stopovers, as shown in Fig. 3—14a. The first leg is due east for till} km; the second leg is southeast (4-5“) for 440m; and the third leg is at 53" south of west, for 550 ion, as shower. What is the plane’s total displacement? APPROACH We follow the steps in the Problem Solving Strategy above SOLUTION 1. Draw a diagram such as Fig. 3—14a. where 131 , fl), and fig represent the three legs of the trip, and fig is the planet total displacement. 2. Choose ates: Ares are also shorten in Fig. 3—14a: n is east, y north. 3. Resolve oomponents: It is imperative to draw a good diagram. The components are drawn in Fig. S-Mb. Instead of drawing all the vectors starting from a common origin, as we did in Fig. 3—13h, here we draw them “tail-totip” style, which is just as valid and may make it easier to see. 4. Calculate the components; fi1:D1x- — +D1cosil" = D1 = $ka D1}. = +D1sintl° = llltui nan” = +D2cos45° = +[44tlltm](tl'ltt't]= +311ltrn ox! = oases: = —[ttatm){o.)=rar —311|m1 li D351)” = -Dgeos53” = -[55[tlrm 0.602]: -331lnn D3]. = - D3 sin 53“ = - (550 km)(ll.l'99]= -439 Iron. We have given a minus sign to each component that in Fig. 3—14]: points in the - x or - y direction. The components are shown in the Table in the margin. 5. Add the components: We add the x eomponents together, and we add the FIGURE 3-" Example 3—3. m y components together to obtain the x and y components of the resultant: Iilflll film} Dr: Du+Dn+D3g = fififlkm+311km—331ltm = IStltlltm rt1 are a = ”or +02? +03? = dim —311km—439km = —tsotm. 91 3‘11 '3“ The x and y components are hill] km and 450 km, and point respectively to W the east and soutthis is one way to give the answer. W 6. Magnitude and direction: WeI can also give the answer as =\/ D§+ [$632+ {—Tjtljzkm = 95ka land = g: = Salim = -1. 25 see = -51°. Thus, the total displacement has magnitude 960 lrrn and points 51“I below the 53 Cl'lAPTER 5 .1: axis (south of east], as was shown] in our original sketch, Fig. 3—l4a. Poslflon given as a function of time. The position of a particle as a function of time is given by i = [(5.0 Ito’s]! + {fiflmfszh‘zli + [[10 cu) — [lilmfsajtali where r is in meters and t is in seconds. [o] What is the particle's displaoenient between :1 = lils and I: = 3.0 s? (b) Determine the particle‘s instantaneous vein-til)" and acceleration as a function of time. [a] Evaluate ir’ and i at t = 3.0 s. APPRDACH For {a}, we find iii" = f; - f1, inserting £1 = ills for finding i=1 i and E1 = 3.133 for ii. For [it], we take derivatiVEs {Eqs 3—9 and 3—11], and for {e} we substitute t = 311s into our results in {b}. sotunou {a} a: :1 = 2.0a, a] = [(5.0 mfsfllfls} + {asmxstttassflt + [{Tflm} — {asmxfigznnflj = [34m]i — {1am}. Similarly, at t; = 311s, i1 = {15m + 54m}i + ("film — s1 mji = {69m}i — (Tami. Thus as = sz-s1 = {fiQm-Mmji+[-T4rn+l?ru]j = [35m}i-(5Trujj. That is, ex = 35 In, and n}- = -5".r' m. (b) To find 1velocity+ we take the derivative of the given it with respect to time+ noting [Appendix 3—2) that d[t1]fdt = 2:, and atrijfat = 3:1: it = g = [5.u mfs + (12 111(51):]? + [a — [9.umfs3}t=]i. The aeceleration is [keeping only two significant figures): it = g = [limifli —[1Smfs3]lti. Thus a; = unit's2 is eonstant; but a}, = —[lflmfs3jt depends linearly on time, increasing in magnitude with time in the negative 3! direction. {c} We substitute t = lfls into the equations we just derived for i and i: a = [Sflmfs + Efimfsfi — (s1 mfsji = {41 mmi — [s1 mm a = (12 maul — [s4 mat-)3“. Their magnitudes at t = 3.[|s are at = “V l[411n,u’s]u2 + {31 nt,.:"s]l2 = 91 refs, and a = "v [12 ruff): + {54 Intel]1 = 55 111,332. Driving off I cliff. A movie stunt driver on a motorcycle speeds horizontally off a Still-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, S‘llflm from the base of the cliff where the cameras are? Ignore air resistance APPROACH We explicitly follow the steps of the Problem Solving Strategy above. SOLUTION 1. and 2. Read, choose the object, and draw a diagram. Our object is the motorcycle and driver, taken as a single unit.The diagram is shown in Fig. 3—23. 3. Choose a coordinate system.We choose the y direction to be positive upward, with the top of the clilf as Jii = t}. The x direction is horizontal with x0 = U at the point where the motorcycle leaves the elilt 4. Choose a time Interval. We choose our time interval to begin [t = 0) just as the motorcycle leaves the cliff top at position In, = 0, )3 = 0; our time interval ends just betore the motorcycle hits the ground below. 5. Examine .r and y motions. to the horizontal (x) direction, the acceleration ex = i], so the velocity is constant. The value of in when the motorcycle reaches the ground is x = +901) In. In the vertical direction, the accelera— tion is the acceleration due to gravity, a}, = -g = -9.30 mfsz. The value of y when the motorcycle reaches the ground is y = —5{].Um. The initial velocity is horizontal and is our unknown, on; the initial vertical velocity is zero, ow = 0. 6. List knowna and unknowna. See the Table in the margin. Note that in addition to not knowing the initial horizontal velocity a”, [which stays constant until landing}, we also do not know the time t when the motorcycle reaches the ground. 7. Apply relevant equations. The motorcycle maintains constant oJr as long as it is in the air. The time it stays in the air is determined by the y motion—when it hits the ground. So we first find the time using the y motion, and then use this time value in the 1' equations Tb find out how long it takes the motorcycle to reach the ground below, we use Eq. 2—12b (Table 3—2) for the vertical (y) direction with ya = 0 and ”I“ = 0: y = g] + 1.5.0! + §riyt1 = n + e + fl—gjtz or y = eat". We solve fort and set y = -5CI.D tn: 2 2 -50.0m r= .i—y = ,il—g = 3.19s. '3' -9.Sflmy’s To calculate the initial velocity1 the: we again use Eq. 2—12b, but this time for the horizontal (x) direction, with a; = D and I.) = I]: a: = x0+trmt+ilaxt2 = t] + amt + t] or .t: = vied. Then x 90.0m = — = = 28.2 , v” r 3.195 "1"5 which is about 100 kmfh (roughly 60 mifh). NOTE In the time interval of the projectile motion, the only acceleration is g in the negative y direction. The acceleration in the x direction is zero. FIGURE 3-13 Example 3—6. a=n=0 x = Sillflm y = —5{i.tint ox = a}. = -g = -§Lfllllmtis2 3}”! = 0 Unknown 1's!) FIGURE 3-!!! Example 3—7. @ansms APPLIED Sp arts J" vy=0atthispoint _________ fi-L-er‘: vyn _! r‘ 7*“ ti=E=-s.l A kicked football. A football is kicked at an angle it.) = 3?.0" with a velocity of 20.0 mls, as shewn in Fig, 3—24. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (e) hov.r far away it hits the ground, ((1) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball. APPROACH This may seem difficult at first because there are so many questions. But we can deal with them one at a time. We take the y direction as positive upward, and treat the x and y motions separately. The total time in the air is again determined by the y motion. The x motion occurs at constant velocity. The y component of velocity varies, being positive (upward) initially. decreasing to zero at the highest point, and then becoming negative as the football falls. SOLUTION W...
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