{
  "index": "1970-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-4. An automobile starts from rest and ends at rest, traversing a distance of one mile in one minute, along a straight road. If a governor prevents the speed of the car from exceeding ninety miles per hour, show that at some time of the traverse the acceleration or deceleration of the car was at least \\( 6.6 \\mathrm{ft} . / \\mathrm{sec} .^{2} \\)",
  "solution": "B-4 Converting units to feet and seconds, we have \\( 0 \\leqq v(t) \\leqq 132 \\) for all \\( t \\in[0,60] \\). Suppose \\( \\left|v^{\\prime}(t)\\right|<6.6 \\) for all \\( t \\in[0,60] \\). Then \\( v(t)=\\int_{0}^{t} v^{\\prime}<6.6 t \\), and \\( v(t)=\\int_{t}^{60}-v^{\\prime}<6.6(60-t) \\) for all \\( t \\in[0,60] \\). Thus\n\\[\n5280=\\int_{0}^{60} v(t) d t<\\int_{0}^{60} \\min \\{6.6 t, 6.6(60-t), 132\\} d t .\n\\]\n\nThis last integral is the area under a trapezoid and equals the value 5280 , which is a contradiction.",
  "vars": [
    "v",
    "t"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "v": "velocity",
        "t": "timevar"
      },
      "question": "B-4. An automobile starts from rest and ends at rest, traversing a distance of one mile in one minute, along a straight road. If a governor prevents the speed of the car from exceeding ninety miles per hour, show that at some time of the traverse the acceleration or deceleration of the car was at least \\( 6.6 \\mathrm{ft} . / \\mathrm{sec} .^{2} \\)",
      "solution": "B-4 Converting units to feet and seconds, we have \\( 0 \\leqq velocity(timevar) \\leqq 132 \\) for all \\( timevar \\in[0,60] \\). Suppose \\( \\left|velocity^{\\prime}(timevar)\\right|<6.6 \\) for all \\( timevar \\in[0,60] \\). Then \\( velocity(timevar)=\\int_{0}^{timevar} velocity^{\\prime}<6.6 timevar \\), and \\( velocity(timevar)=\\int_{timevar}^{60}-velocity^{\\prime}<6.6(60-timevar) \\) for all \\( timevar \\in[0,60] \\). Thus\n\\[\n5280=\\int_{0}^{60} velocity(timevar) d timevar<\\int_{0}^{60} \\min \\{6.6 timevar, 6.6(60-timevar), 132\\} d timevar .\n\\]\n\nThis last integral is the area under a trapezoid and equals the value 5280 , which is a contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "v": "turnstile",
        "t": "marigold"
      },
      "question": "B-4. An automobile starts from rest and ends at rest, traversing a distance of one mile in one minute, along a straight road. If a governor prevents the speed of the car from exceeding ninety miles per hour, show that at some time of the traverse the acceleration or deceleration of the car was at least \\( 6.6 \\mathrm{ft} . / \\mathrm{sec} .^{2} \\)",
      "solution": "B-4 Converting units to feet and seconds, we have \\( 0 \\leqq turnstile(marigold) \\leqq 132 \\) for all \\( marigold \\in[0,60] \\). Suppose \\( \\left| turnstile^{\\prime}(marigold) \\right|<6.6 \\) for all \\( marigold \\in[0,60] \\). Then \\( turnstile(marigold)=\\int_{0}^{marigold} turnstile^{\\prime}<6.6 marigold \\), and \\( turnstile(marigold)=\\int_{marigold}^{60}-turnstile^{\\prime}<6.6(60-marigold) \\) for all \\( marigold \\in[0,60] \\). Thus\n\\[\n5280=\\int_{0}^{60} turnstile(marigold) d marigold<\\int_{0}^{60} \\min \\{6.6 marigold, 6.6(60-marigold), 132\\} d marigold .\n\\]\n\nThis last integral is the area under a trapezoid and equals the value 5280 , which is a contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "v": "immobility",
        "t": "timelessness"
      },
      "question": "B-4. An automobile starts from rest and ends at rest, traversing a distance of one mile in one minute, along a straight road. If a governor prevents the speed of the car from exceeding ninety miles per hour, show that at some time of the traverse the acceleration or deceleration of the car was at least \\( 6.6 \\mathrm{ft} . / \\mathrm{sec} .^{2} \\)",
      "solution": "B-4 Converting units to feet and seconds, we have \\( 0 \\leqq immobility(timelessness) \\leqq 132 \\) for all \\( timelessness \\in[0,60] \\). Suppose \\( \\left|immobility^{\\prime}(timelessness)\\right|<6.6 \\) for all \\( timelessness \\in[0,60] \\). Then \\( immobility(timelessness)=\\int_{0}^{timelessness} immobility^{\\prime}<6.6\\, timelessness \\), and \\( immobility(timelessness)=\\int_{timelessness}^{60}-immobility^{\\prime}<6.6(60-timelessness) \\) for all \\( timelessness \\in[0,60] \\). Thus\n\\[\n5280=\\int_{0}^{60} immobility(timelessness) \\, d timelessness<\\int_{0}^{60} \\min \\{6.6\\, timelessness, 6.6(60-timelessness), 132\\} \\, d timelessness .\n\\]\n\nThis last integral is the area under a trapezoid and equals the value 5280 , which is a contradiction."
    },
    "garbled_string": {
      "map": {
        "v": "plkqsjmo",
        "t": "xqzpjrlu"
      },
      "question": "B-4. An automobile starts from rest and ends at rest, traversing a distance of one mile in one minute, along a straight road. If a governor prevents the speed of the car from exceeding ninety miles per hour, show that at some time of the traverse the acceleration or deceleration of the car was at least \\( 6.6 \\mathrm{ft} . / \\mathrm{sec} .^{2} \\)",
      "solution": "B-4 Converting units to feet and seconds, we have \\( 0 \\leqq plkqsjmo(xqzpjrlu) \\leqq 132 \\) for all \\( xqzpjrlu \\in[0,60] \\). Suppose \\( \\left|plkqsjmo^{\\prime}(xqzpjrlu)\\right|<6.6 \\) for all \\( xqzpjrlu \\in[0,60] \\). Then \\( plkqsjmo(xqzpjrlu)=\\int_{0}^{xqzpjrlu} plkqsjmo^{\\prime}<6.6 xqzpjrlu \\), and \\( plkqsjmo(xqzpjrlu)=\\int_{xqzpjrlu}^{60}-plkqsjmo^{\\prime}<6.6(60-xqzpjrlu) \\) for all \\( xqzpjrlu \\in[0,60] \\). Thus\n\\[\n5280=\\int_{0}^{60} plkqsjmo(xqzpjrlu) d xqzpjrlu<\\int_{0}^{60} \\min \\{6.6 xqzpjrlu, 6.6(60-xqzpjrlu), 132\\} d xqzpjrlu .\n\\]\n\nThis last integral is the area under a trapezoid and equals the value 5280 , which is a contradiction."
    },
    "kernel_variant": {
      "question": "Consider again the prototype maglev capsule that travels along a perfectly straight test–track.  \n\nDuring the run it  \n• starts from rest at platform $A$ at $t=0$,  \n• returns to rest at platform $B$ at $t=T$,  \n\nwith  \n\n\\[\nT = 90.0\\ {\\rm s},\\qquad  \nD = 2\\,850\\ {\\rm m}.\n\\]\n\nThroughout the motion the following {\\it independent safety rules} are simultaneously enforced.\n\n1.\\;{\\bf Time–dependent speed ceilings}\n\\[\n\\begin{array}{ll}\n0\\le t\\le 30.0\\;{\\rm s}: & v(t)\\le 25.0\\ {\\rm m\\,s^{-1}},\\\\[4pt]\n30.0\\;{\\rm s}<t\\le 60.0\\;{\\rm s}: & v(t)\\le 45.0\\ {\\rm m\\,s^{-1}},\\\\[4pt]\n60.0\\;{\\rm s}<t\\le 90.0\\;{\\rm s}: & v(t)\\le 25.0\\ {\\rm m\\,s^{-1}}.\n\\end{array}\n\\]\n\n2.\\;{\\bf Global jerk bound}  \nThe velocity is twice continuously differentiable and its second derivative (the jerk) satisfies  \n\\[\n\\lvert j(t)\\rvert=\\lvert v''(t)\\rvert\\le J,\n\\qquad J = 3.50\\ {\\rm m\\,s^{-3}}.\n\\]\n\n3.\\;{\\bf Straight track}  \nAll kinematical quantities are purely tangential; hence $v(t)\\ge 0$ and\n$a(t)=v'(t)$ is a signed scalar.\n\nShow that, no matter how ingeniously the run is programmed, there exists an\ninstant $t^{\\ast}\\in(0,T)$ at which the magnitude of the acceleration is at least  \n\n\\[\n\\boxed{\\;\n\\lvert a(t^{\\ast})\\rvert \\,\\ge\\, a_{\\min},\n\\qquad a_{\\min}= 3.00\\ {\\rm m\\,s^{-2}}\\; }.\n\\]\n\nIn other words, {\\it some} point of the journey must experience an\nacceleration (or deceleration) of at least $3.00\\ {\\rm m\\,s^{-2}}$.",
      "solution": "We argue by contradiction.  Denote  \n\n\\[\na(t)=v'(t),\\qquad j(t)=v''(t),\\qquad 0\\le t\\le T=90\\ {\\rm s}.\n\\]\n\nThroughout the proof we assume  \n\n\\[\n\\lvert a(t)\\rvert<A,\\qquad A:=3.00\\ {\\rm m\\,s^{-2}}\n\\qquad\\forall t\\in[0,T]\n\\tag{H}\n\\]\nand show that under the three safety rules the travelled distance\ncannot reach $D$.  This contradicts the data and will prove the theorem.\n\nStep 1 – Universal forward and backward envelopes  \nBecause of the acceleration bound, for $0\\le t\\le T$\n\\[\nv(t)=\\int_{0}^{t}a(\\tau)\\,d\\tau\\le A\\,t,\n\\qquad \nv(t)=\\int_{t}^{T}(-a)(\\tau)\\,d\\tau\\le A\\,(T-t).\n\\tag{1.1}\n\\]\nCombining (1.1) with the prescribed speed ceilings we introduce the single upper envelope  \n\\[\nw(t):=\\min\\bigl\\{A\\,t,\\,A\\,(T-t),\\,v_{\\rm lim}(t)\\bigr\\},\n\\tag{1.2}\n\\]\nwhere\n\\[\nv_{\\rm lim}(t)=\n\\begin{cases}\n25,&0\\le t\\le 30,\\\\[4pt]\n45,&30<t\\le 60,\\\\[4pt]\n25,&60<t\\le 90.\n\\end{cases}\n\\]\nClearly $v(t)\\le w(t)$ for the entire run.\n\nStep 2 – The first and last $30\\ \\!{\\rm s}$ windows  \nFor $0\\le t\\le30$ we have $w(t)=\\min\\{A\\,t,25\\}$.\nWith $t_{0}:=25/A=8.333\\ {\\rm s}$,\n\\[\n\\int_{0}^{30}w(t)\\,dt=\n\\int_{0}^{t_{0}}A\\,t\\,dt+\\int_{t_{0}}^{30}25\\,dt\n=\\frac{A}{2}t_{0}^{2}+25\\bigl(30-t_{0}\\bigr)\n=645.8\\ {\\rm m}.\n\\]\nBecause $w(t)$ is symmetric with respect to $t=T/2$, the last\nwindow contributes the same amount, whence\n\\[\nd_{1}:=\\int_{0}^{30}v(t)\\,dt\\le645.8\\ {\\rm m},\\qquad\nd_{3}:=\\int_{60}^{90}v(t)\\,dt\\le645.8\\ {\\rm m}.\n\\tag{2.1}\n\\]\n\nStep 3 – Minimal duration of a $25\\ {\\rm m\\,s^{-1}}\\!\\to\\!45\\ {\\rm m\\,s^{-1}}$ ramp  \n\nBefore the middle speed ceiling of $45\\ {\\rm m\\,s^{-1}}$ can be used, the\nspeed must rise from $25$ to $45$ and, later, return to $25$.\nHow fast can a single $+20\\ {\\rm m\\,s^{-1}}$ change be executed if\n$\\lvert a\\rvert<A$ and $\\lvert j\\rvert\\le J$?\n\nOptimal–control fact.                                                              \nA classical application of Pontryagin’s Maximum Principle to a\none–dimensional system $\\dot v=a$, $\\dot a=j$ with control $j$ and bounds\n$\\lvert j\\rvert\\le J$, $\\lvert a\\rvert\\le A$\nshows that the {\\it time–optimal} way to realise a given velocity\nincrement $\\Delta v$ while starting and ending with $a=0$ is a\n{\\it bang–bang–jerk} curve  \n\n$j=+J$ for $0<t<t_{1}$,\\quad\n$j=0$ for $t_{1}<t<t_{1}+t_{2}$,\\quad\n$j=-J$ for $t_{1}+t_{2}<t<t_{1}+t_{2}+t_{1}$,\n\nwhere $t_{1}=A/J$ and $t_{2}=(\\Delta v-A^{2}/J)/A$ if\n$\\Delta v\\ge A^{2}/J$ (otherwise one uses a symmetric triangular\nprofile with $t_{2}=0$).  Any other admissible\nprofile takes at least this long.\n\nFor the present data  \n\\[\n\\Delta v = 20\\ {\\rm m\\,s^{-1}},\\quad\nA = 3.00\\ {\\rm m\\,s^{-2}},\\quad\nJ = 3.50\\ {\\rm m\\,s^{-3}},\\quad\n\\frac{A^{2}}{J}=2.571\\,{\\rm m\\,s^{-1}}<\\Delta v,\n\\]\nso the rectangular–in–$a$ case applies, giving\n\\[\nt_{1}=\\frac{A}{J}=0.8571\\ {\\rm s},\\qquad\nt_{2}=\\frac{\\Delta v-A^{2}/J}{A}=5.810\\ {\\rm s},\n\\]\nhence the minimal rise time is\n\\[\nt_{\\rm ramp}^{\\min}=2t_{1}+t_{2}=7.524\\ {\\rm s}.\n\\tag{3.1}\n\\]\nExactly the same duration is needed for the subsequent\n$45\\to25\\ {\\rm m\\,s^{-1}}$ fall.\n\nRemark.  \nIf an ingenious profile were able to shorten either ramp further (it\ncannot, but let us pretend it could), the extra distance gained during\nthe prolonged $45\\ {\\rm m\\,s^{-1}}$ cruise would be exactly offset by\nthe shorter distance accumulated inside the ramp itself (see Step 4\nbelow).  Thus the forthcoming distance bound would remain unchanged,\nso our final contradiction does {\\it not} rely on the precise value\n$7.524\\ {\\rm s}$.\n\nStep 4 – Distance that can be covered during one ramp  \nDuring the first $t_{s}:=20/A=6.667\\ {\\rm s}$ of the rise the envelope\n(1.2) reads $w(t)=25+A\\,t$; afterwards it is capped by $45$.  Therefore\n\\[\n\\begin{aligned}\nd_{\\rm ramp}\n&\\le\\int_{0}^{t_{s}}\\bigl(25+A\\,t\\bigr)\\,dt\n      +\\int_{t_{s}}^{t_{\\rm ramp}^{\\min}}45\\,dt\\\\\n&=25\\,t_{s}+\\frac{A}{2}t_{s}^{2}+45\\bigl(t_{\\rm ramp}^{\\min}-t_{s}\\bigr)\n=271.9\\ {\\rm m}\\quad(\\text{rounded up to }272\\ {\\rm m}).\n\\end{aligned}\n\\tag{4.1}\n\\]\nTwo such ramps are required, so\n\\[\nd_{\\rm ramps}:=\\text{distance during both ramps}\\le543.8\\ {\\rm m}.\n\\tag{4.2}\n\\]\n\nStep 5 – The remaining {\\it cruise} time in the middle window  \nThe $30\\ {\\rm s}$ middle window leaves\n\\[\nt_{\\rm cruise}=30-2t_{\\rm ramp}^{\\min}=14.95\\ {\\rm s}\n\\tag{5.1}\n\\]\navailable for cruising at most at $45\\ {\\rm m\\,s^{-1}}$, whence\n\\[\nd_{\\rm cruise}\\le45\\,t_{\\rm cruise}=672.8\\ {\\rm m}.\n\\tag{5.2}\n\\]\nConsequently\n\\[\nd_{2}:=\\int_{30}^{60}v(t)\\,dt\n\\le d_{\\rm ramps}+d_{\\rm cruise}\n\\le543.8+672.8=1\\,216.6\\ {\\rm m}.\n\\tag{5.3}\n\\]\n\nStep 6 – Global upper bound on the travelled distance  \nCollecting (2.1) and (5.3) we obtain\n\\[\n\\int_{0}^{T}v(t)\\,dt\n\\le d_{1}+d_{2}+d_{3}\n\\le645.8+1\\,216.6+645.8\n=2\\,508.2\\ {\\rm m}\\;<\\;D.\n\\]\n\nStep 7 – Contradiction and conclusion  \nUnder hypothesis (H) the capsule can cover at most\n$2\\,508\\ {\\rm m}$, far short of the required $2\\,850\\ {\\rm m}$.\nTherefore (H) is impossible, i.e.  \n\\[\n\\exists\\,t^{\\ast}\\in(0,90)\\quad\\text{such that}\\quad\n\\lvert a(t^{\\ast})\\rvert\\ge3.00\\ {\\rm m\\,s^{-2}}.\n\\]\nThe claim is proved.\n\n{\\small\nNumerical optimisation (not required for the proof) confirms that the\nsmallest attainable peak acceleration compatible with all safety rules\nis about $5.4\\ {\\rm m\\,s^{-2}}$; thus the bound $3.00\\ {\\rm m\\,s^{-2}}$\nis non–trivial yet not sharp.\n}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.594761",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional constraints  \n   • The original and kernel problems involved only a speed cap; the\n     enhanced variant also keeps the jerk bounded, which forces a\n     completely new quadratic upper bound for the speed near the end\n     points.  \n   • The speed cap itself is piece-wise constant with three different\n     windows instead of a single uniform limit.\n\n2. Higher technical content  \n   • One must combine five separate upper envelopes (two quadratic, two\n     linear and one rectangular) and integrate the pointwise minimum.  \n   • Locating the intersection abscissae and deciding which pieces are\n     active for a given tentative A is already a non-trivial logical\n     step and cannot be handled by simple pattern matching.\n\n3. Deeper theory  \n   • The proof uses jerk–limited kinematics, a topic that normally\n     appears only in advanced optimal-control courses.  \n   • Sharpness is justified by an explicit construction that saturates\n     simultaneously the jerk bound, the acceleration bound and the\n     piece-wise speed bound—an optimisation problem of higher order than\n     the one in the original statement.\n\n4. Significantly more work  \n   • The original solution needs one trapezoidal area calculation; the\n     present one requires analysing three regimes and integrating five\n     different elementary functions over several sub-intervals.  \n   • The appearance of quadratic envelopes forces careful bookkeeping of\n     intersection points and inevitably lengthens the computation.\n\nFor all these reasons the enhanced kernel variant is markedly harder\nthan both the original B-4 problem and the easier “tram’’ kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Consider again the prototype maglev capsule that travels along a perfectly straight test–track.  \nDuring the run it\n\n• starts from rest at platform $A$ at $t=0$,  \n• returns to rest at platform $B$ at $t=T$,\n\nwith  \n\n\\[\nT = 90.0\\ {\\rm s},\\qquad  \nD = 2\\,850\\ {\\rm m}.\n\\]\n\nThroughout the motion the following {\\it independent safety rules} are simultaneously enforced.\n\n1. (Time–dependent speed ceilings)\n\\[\n\\begin{array}{ll}\n0\\le t\\le 30.0\\;{\\rm s}: & v(t)\\le 25.0\\ {\\rm m\\,s^{-1}},\\\\[4pt]\n30.0\\;{\\rm s}<t\\le 60.0\\;{\\rm s}: & v(t)\\le 45.0\\ {\\rm m\\,s^{-1}},\\\\[4pt]\n60.0\\;{\\rm s}<t\\le 90.0\\;{\\rm s}: & v(t)\\le 25.0\\ {\\rm m\\,s^{-1}}.\n\\end{array}\n\\]\n\n2. (Global jerk bound)  \nThe velocity is twice continuously differentiable and its second derivative (the\njerk) satisfies\n\\[\n\\lvert j(t)\\rvert=\\lvert v''(t)\\rvert\\le J,\n\\qquad J = 3.50\\ {\\rm m\\,s^{-3}}.\n\\]\n\n3. (Straight track)  \nAll kinematical quantities are purely tangential; hence $v(t)\\ge 0$ and\n$a(t)=v'(t)$ is a signed scalar.\n\nShow that, no matter how ingeniously the run is programmed, there exists an\ninstant $t^{\\ast}\\in(0,T)$ at which the magnitude of the acceleration is at least  \n\n\\[\n\\boxed{\\;\n\\lvert a(t^{\\ast})\\rvert \\,\\ge\\, a_{\\min},\n\\qquad a_{\\min}= 3.00\\ {\\rm m\\,s^{-2}}\\; }.\n\\]\n\nIn other words, {\\it some} point of the journey must experience an\nacceleration (or deceleration) of at least $3.00\\ {\\rm m\\,s^{-2}}$.",
      "solution": "We argue by contradiction.  Denote  \n\n\\[\na(t)=v'(t),\\qquad j(t)=v''(t),\\qquad 0\\le t\\le T=90\\ {\\rm s}.\n\\]\n\nThroughout the proof we assume  \n\n\\[\n\\lvert a(t)\\rvert<A,\\qquad A:=3.00\\ {\\rm m\\,s^{-2}}\n\\qquad\\forall t\\in[0,T]\n\\tag{H}\n\\]\nand show that under the three safety rules the travelled distance\ncannot reach $D$.  This contradicts the data and will prove the theorem.\n\nStep 1 – Universal forward and backward envelopes  \nBecause of the jerk bound and the unknown initial acceleration, for every\n$t\\ge0$\n\\[\na(t)\\le a(0)+Jt< A+Jt.\n\\]\nConsequently  \n\\[\na(t)\\le\\min\\{A,\\,a(0)+Jt\\}\\le A \\qquad(0\\le t\\le T).\n\\]\nIntegrating once gives the {\\it forward} velocity bound  \n\\[\nv(t)=\\int_{0}^{t}a(\\tau)\\,d\\tau\\le\\int_{0}^{t}A\\,d\\tau=A\\,t.\n\\tag{1.1}\n\\]\nA completely analogous argument from the right end--point yields the\n{\\it backward} bound  \n\\[\nv(t)=\\int_{t}^{T}(-a)(\\tau)\\,d\\tau\\le A\\,(T-t).\n\\tag{1.2}\n\\]\nBoth estimates are valid irrespective of the (unknown) values of\n$a(0)$ and $a(T)$.\n\nCombining (1.1), (1.2) and the prescribed speed ceilings we introduce\n\\[\nw(t):=\\min\\bigl\\{A\\,t,\\,A\\,(T-t),\\,v_{\\rm lim}(t)\\bigr\\},\n\\tag{1.3}\n\\]\nwhere\n\\[\nv_{\\rm lim}(t)=\n\\begin{cases}\n25,&0\\le t\\le 30,\\\\[4pt]\n45,&30<t\\le 60,\\\\[4pt]\n25,&60<t\\le 90.\n\\end{cases}\n\\]\n\nStep 2 – The first and last $30\\,{\\rm s}$ windows  \nFor $0\\le t\\le30$ we have $w(t)=\\min\\{A\\,t,25\\}$.\nWith $t_{0}:=25/A=8.333\\ {\\rm s}$,\n\\[\n\\int_{0}^{30}w(t)\\,dt=\n\\int_{0}^{t_{0}}A\\,t\\,dt+\\int_{t_{0}}^{30}25\\,dt\n=\\frac{A}{2}t_{0}^{2}+25\\,(30-t_{0})\\;=\\;645.8\\ {\\rm m}.\n\\]\nBecause $w(t)$ is symmetric with respect to the midpoint\n$t=\\tfrac{T}{2}$, the last thirty–second window contributes the same\namount.  Thus\n\\[\nd_{1}:=\\int_{0}^{30}v(t)\\,dt\\le645.8\\ {\\rm m},\\qquad\nd_{3}:=\\int_{60}^{90}v(t)\\,dt\\le645.8\\ {\\rm m}.\n\\tag{2.1}\n\\]\n\nStep 3 – Minimal duration of a single $25\\rightarrow45$ velocity ramp  \nBetween $t=30$ and $t=60$ the speed must rise from $25$ to $45$ and,\nlater, return to $25$.  Consider the {\\it rise}.\nBecause $\\lvert a\\rvert<A$ and $\\lvert j\\rvert\\le J$, the most\ntime–efficient profile is\n\n• $j=+J$ for $t_{1}:=A/J=0.8571\\ {\\rm s}$ (bringing $a$ from $0$ to $A$);  \n• $a=A$ for  \n\\[\nt_{2}:=\\frac{20-A^{2}/J}{A}=5.810\\ {\\rm s};\n\\]\n• $j=-J$ for another $t_{1}$ (reducing $a$ back to $0$).\n\nHence the duration of one ramp is  \n\\[\nt_{\\rm ramp}=2t_{1}+t_{2}=7.524\\ {\\rm s}.\n\\tag{3.1}\n\\]\n\nStep 4 – Distance that can be covered during one ramp  \nDuring the first $t_{s}:=20/A=6.667\\ {\\rm s}$ of the ramp the speed is\nbounded by $v(t)\\le25+A\\,t$; afterwards it is capped by $45$.  Therefore\n\n\\[\n\\begin{aligned}\nd_{\\rm ramp}\n&\\le\\int_{0}^{t_{s}}(25+A\\,t)\\,dt+\\int_{t_{s}}^{t_{\\rm ramp}}45\\,dt\\\\\n&=25t_{s}+\\frac{A}{2}t_{s}^{2}+45\\bigl(t_{\\rm ramp}-t_{s}\\bigr)\\\\\n&=271.9\\ {\\rm m}\\quad(\\text{rounded up to }272\\ {\\rm m}).\n\\end{aligned}\n\\tag{4.1}\n\\]\nTwo such ramps are required, so\n\n\\[\nd_{\\rm ramps}:=\\text{distance during both ramps}\\le543.8\\ {\\rm m}.\n\\tag{4.2}\n\\]\n\nStep 5 – The remaining {\\it cruise} time in the middle window  \nThe $30\\ {\\rm s}$ middle window leaves\n\n\\[\nt_{\\rm cruise}=30-2t_{\\rm ramp}=14.95\\ {\\rm s}\n\\tag{5.1}\n\\]\navailable for cruising at most at $45\\ {\\rm m\\,s^{-1}}$, whence\n\n\\[\nd_{\\rm cruise}\\le45\\,t_{\\rm cruise}=672.8\\ {\\rm m}.\n\\tag{5.2}\n\\]\n\nConsequently\n\n\\[\nd_{2}:=\\int_{30}^{60}v(t)\\,dt\n\\le d_{\\rm ramps}+d_{\\rm cruise}\n\\le543.8+672.8=1\\,216.6\\ {\\rm m}.\n\\tag{5.3}\n\\]\n\nStep 6 – Global upper bound on the travelled distance  \nCollecting (2.1) and (5.3) we obtain\n\n\\[\n\\int_{0}^{T}v(t)\\,dt\n\\le d_{1}+d_{2}+d_{3}\n\\le645.8+1\\,216.6+645.8\n=2\\,508.2\\ {\\rm m}\\;<\\;D.\n\\]\n\nStep 7 – Contradiction and conclusion  \nUnder hypothesis (H) the capsule can cover at most\n$2\\,508\\ {\\rm m}$, well short of the required $2\\,850\\ {\\rm m}$.\nTherefore (H) is impossible, i.e.  \n\\[\n\\exists\\,t^{\\ast}\\in(0,90)\\quad\\text{such that}\\quad\n\\lvert a(t^{\\ast})\\rvert\\ge3.00\\ {\\rm m\\,s^{-2}}.\n\\]\nThe claim is proved.\n\n{\\small\nNumerical optimisation (not required for the proof) shows that the\nminimal attainable peak acceleration compatible with all safety rules is\nabout $5.4\\ {\\rm m\\,s^{-2}}$; hence the bound $3.00\\ {\\rm m\\,s^{-2}}$\nis non–trivial yet still improvable.\n}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.476313",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional constraints  \n   • The original and kernel problems involved only a speed cap; the\n     enhanced variant also keeps the jerk bounded, which forces a\n     completely new quadratic upper bound for the speed near the end\n     points.  \n   • The speed cap itself is piece-wise constant with three different\n     windows instead of a single uniform limit.\n\n2. Higher technical content  \n   • One must combine five separate upper envelopes (two quadratic, two\n     linear and one rectangular) and integrate the pointwise minimum.  \n   • Locating the intersection abscissae and deciding which pieces are\n     active for a given tentative A is already a non-trivial logical\n     step and cannot be handled by simple pattern matching.\n\n3. Deeper theory  \n   • The proof uses jerk–limited kinematics, a topic that normally\n     appears only in advanced optimal-control courses.  \n   • Sharpness is justified by an explicit construction that saturates\n     simultaneously the jerk bound, the acceleration bound and the\n     piece-wise speed bound—an optimisation problem of higher order than\n     the one in the original statement.\n\n4. Significantly more work  \n   • The original solution needs one trapezoidal area calculation; the\n     present one requires analysing three regimes and integrating five\n     different elementary functions over several sub-intervals.  \n   • The appearance of quadratic envelopes forces careful bookkeeping of\n     intersection points and inevitably lengthens the computation.\n\nFor all these reasons the enhanced kernel variant is markedly harder\nthan both the original B-4 problem and the easier “tram’’ kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}