{
  "index": "1972-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-2. A particle moving on a straight line starts from rest and attains a velocity \\( v_{0} \\) after traversing a distance \\( s_{0} \\). If the motion is such that the acceleration was never increasing, find the maximum time for the traverse.",
  "solution": "B-2 We take \\( v_{0} \\) as positive (see Comment) and consider the graph of \\( v \\) as a function of \\( t \\) (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration \\( a=d v / d t \\) does not increase.\n\nFig. 1\n\nLet \\( t_{0} \\) be the time of the traverse. Then \\( v\\left(t_{0}\\right)=v_{0} \\). The distance \\( s_{0} \\) is represented by the area bounded by the curve \\( v=v(t) \\), the \\( t \\)-axis, and the line \\( t=t_{0} \\). The area of the right triangle with vertices at \\( (0,0),\\left(t_{0}, 0\\right) \\) and \\( \\left(t_{0}, v_{0}\\right) \\) has area less than or equal to \\( s \\). Thus \\( \\frac{1}{2} v_{0} t_{0} \\leqq s_{0} \\) or\n\\[\nt_{0} \\leqq \\frac{2 s_{0}}{v_{0}} .\n\\]\n\nEquality is possible and gives the maximum value of \\( t_{0} \\) (for given \\( s_{0} \\) and \\( v_{0} \\) ) when the graph of \\( v(t) \\) is the straight line \\( v(t)=\\left(v_{0} / t_{0}\\right) t=\\left(v_{0}^{2} / 2 s_{0}\\right) t \\).\n\nComment: If \\( v_{0} \\) is zero or negative, there is no maximum time \\( t_{0} \\) for the traverse. In the case \\( v_{0}=0 \\) the equation of motion\n\\[\nS=s_{0}\\left[3\\left(t / t_{0}\\right)^{2}-3\\left(t / t_{0}\\right)^{3}\\right], \\quad 0 \\leqq t \\leqq t_{0}\n\\]\nsatisfies the conditions of the problem for any \\( t_{0}>0 \\).",
  "vars": [
    "v",
    "a",
    "t",
    "t_0",
    "S",
    "s"
  ],
  "params": [
    "v_0",
    "s_0"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "v": "velocity",
        "a": "accelr",
        "t": "timevar",
        "t_0": "traversetime",
        "S": "displcap",
        "s": "displlow",
        "v_0": "initveloc",
        "s_0": "initdispl"
      },
      "question": "B-2. A particle moving on a straight line starts from rest and attains a velocity \\( initveloc \\) after traversing a distance \\( initdispl \\). If the motion is such that the acceleration was never increasing, find the maximum time for the traverse.",
      "solution": "B-2 We take \\( initveloc \\) as positive (see Comment) and consider the graph of \\( velocity \\) as a function of \\( timevar \\) (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration \\( accelr = d\\,velocity / d\\,timevar \\) does not increase.\n\nFig. 1\n\nLet \\( traversetime \\) be the time of the traverse. Then \\( velocity\\!\\left(traversetime\\right)=initveloc \\). The distance \\( initdispl \\) is represented by the area bounded by the curve \\( velocity = velocity(timevar) \\), the \\( timevar \\)-axis, and the line \\( timevar = traversetime \\). The area of the right triangle with vertices at \\( (0,0),\\,(traversetime,0) \\) and \\( (traversetime,\\,initveloc) \\) has area less than or equal to \\( displlow \\). Thus \\( \\frac{1}{2}\\, initveloc\\, traversetime \\leqq initdispl \\) or\n\\[\ntraversetime \\leqq \\frac{2\\, initdispl}{initveloc} .\n\\]\n\nEquality is possible and gives the maximum value of \\( traversetime \\) (for given \\( initdispl \\) and \\( initveloc \\) ) when the graph of \\( velocity(timevar) \\) is the straight line \\( velocity(timevar)=\\left(initveloc / traversetime\\right) timevar = \\left(initveloc^{2} / 2\\,initdispl\\right) timevar \\).\n\nComment: If \\( initveloc \\) is zero or negative, there is no maximum time \\( traversetime \\) for the traverse. In the case \\( initveloc = 0 \\) the equation of motion\n\\[\ndisplcap = initdispl\\Bigl[3\\bigl(timevar / traversetime\\bigr)^{2} - 3\\bigl(timevar / traversetime\\bigr)^{3}\\Bigr], \\quad 0 \\leqq timevar \\leqq traversetime\n\\]\nsatisfies the conditions of the problem for any \\( traversetime > 0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "v": "pineapple",
        "a": "suitcase",
        "t": "lanterns",
        "t_0": "harmonica",
        "S": "margarine",
        "s": "toothbrush",
        "v_0": "blueberry",
        "s_0": "chandelier"
      },
      "question": "B-2. A particle moving on a straight line starts from rest and attains a velocity \\( blueberry \\) after traversing a distance \\( chandelier \\). If the motion is such that the acceleration was never increasing, find the maximum time for the traverse.",
      "solution": "B-2 We take \\( blueberry \\) as positive (see Comment) and consider the graph of \\( pineapple \\) as a function of \\( lanterns \\) (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration \\( suitcase=d pineapple / d lanterns \\) does not increase.\n\nFig. 1\n\nLet \\( harmonica \\) be the time of the traverse. Then \\( pineapple\\left(harmonica\\right)=blueberry \\). The distance \\( chandelier \\) is represented by the area bounded by the curve \\( pineapple=pineapple(lanterns) \\), the \\( lanterns \\)-axis, and the line \\( lanterns=harmonica \\). The area of the right triangle with vertices at \\( (0,0),\\left(harmonica, 0\\right) \\) and \\( \\left(harmonica, blueberry\\right) \\) has area less than or equal to \\( toothbrush \\). Thus \\( \\frac{1}{2} blueberry harmonica \\leqq chandelier \\) or\n\\[\nharmonica \\leqq \\frac{2 chandelier}{blueberry} .\n\\]\n\nEquality is possible and gives the maximum value of \\( harmonica \\) (for given \\( chandelier \\) and \\( blueberry \\) ) when the graph of \\( pineapple(lanterns) \\) is the straight line \\( pineapple(lanterns)=\\left(blueberry / harmonica\\right) lanterns=\\left(blueberry^{2} / 2 chandelier\\right) lanterns \\).\n\nComment: If \\( blueberry \\) is zero or negative, there is no maximum time \\( harmonica \\) for the traverse. In the case \\( blueberry=0 \\) the equation of motion\n\\[\nmargarine=chandelier\\left[3\\left(lanterns / harmonica\\right)^{2}-3\\left(lanterns / harmonica\\right)^{3}\\right], \\quad 0 \\leqq lanterns \\leqq harmonica\n\\]\nsatisfies the conditions of the problem for any \\( harmonica>0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "v": "stillness",
        "a": "stagnation",
        "t": "timelessness",
        "t_0": "timelesszero",
        "S": "restdistance",
        "s": "standstill",
        "v_0": "restfulspeed",
        "s_0": "restdistanceinit"
      },
      "question": "B-2. A particle moving on a straight line starts from rest and attains a velocity \\( restfulspeed \\) after traversing a distance \\( restdistanceinit \\). If the motion is such that the acceleration was never increasing, find the maximum time for the traverse.",
      "solution": "B-2 We take \\( restfulspeed \\) as positive (see Comment) and consider the graph of \\( stillness \\) as a function of \\( timelessness \\) (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration \\( stagnation=d stillness / d timelessness \\) does not increase.\n\nFig. 1\n\nLet \\( timelesszero \\) be the time of the traverse. Then \\( stillness\\left(timelesszero\\right)=restfulspeed \\). The distance \\( restdistanceinit \\) is represented by the area bounded by the curve \\( stillness=stillness(timelessness) \\), the \\( timelessness \\)-axis, and the line \\( timelessness=timelesszero \\). The area of the right triangle with vertices at \\( (0,0),\\left(timelesszero, 0\\right) \\) and \\( \\left(timelesszero, restfulspeed\\right) \\) has area less than or equal to \\( standstill \\). Thus \\( \\frac{1}{2} restfulspeed\\; timelesszero \\leqq restdistanceinit \\) or\n\\[\ntimelesszero \\leqq \\frac{2\\; restdistanceinit}{restfulspeed} .\n\\]\n\nEquality is possible and gives the maximum value of \\( timelesszero \\) (for given \\( restdistanceinit \\) and \\( restfulspeed \\) ) when the graph of \\( stillness(timelessness) \\) is the straight line \\( stillness(timelessness)=\\left(restfulspeed / timelesszero\\right) timelessness=\\left(restfulspeed^{2} / 2\\; restdistanceinit\\right) timelessness \\).\n\nComment: If \\( restfulspeed \\) is zero or negative, there is no maximum time \\( timelesszero \\) for the traverse. In the case \\( restfulspeed=0 \\) the equation of motion\n\\[\nrestdistance = restdistanceinit\\left[3\\left(timelessness / timelesszero\\right)^{2}-3\\left(timelessness / timelesszero\\right)^{3}\\right], \\quad 0 \\leqq timelessness \\leqq timelesszero\n\\]\nsatisfies the conditions of the problem for any \\( timelesszero>0 \\)."
    },
    "garbled_string": {
      "map": {
        "v": "qzxwvtnp",
        "a": "hjgrksla",
        "t": "mbvcxzpo",
        "t_0": "lkjhgfds",
        "S": "poiuytre",
        "s": "mnbvcxza",
        "v_0": "asdfghjk",
        "s_0": "zxcvbnml"
      },
      "question": "B-2. A particle moving on a straight line starts from rest and attains a velocity \\( asdfghjk \\) after traversing a distance \\( zxcvbnml \\). If the motion is such that the acceleration was never increasing, find the maximum time for the traverse.",
      "solution": "B-2 We take \\( asdfghjk \\) as positive (see Comment) and consider the graph of \\( qzxwvtnp \\) as a function of \\( mbvcxzpo \\) (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration \\( hjgrksla=d qzxwvtnp / d mbvcxzpo \\) does not increase.\n\nFig. 1\n\nLet \\( lkjhgfds \\) be the time of the traverse. Then \\( qzxwvtnp\\left(lkjhgfds\\right)=asdfghjk \\). The distance \\( zxcvbnml \\) is represented by the area bounded by the curve \\( qzxwvtnp=qzxwvtnp(mbvcxzpo) \\), the \\( mbvcxzpo \\)-axis, and the line \\( mbvcxzpo=lkjhgfds \\). The area of the right triangle with vertices at \\( (0,0),\\left(lkjhgfds, 0\\right) \\) and \\( \\left(lkjhgfds, asdfghjk\\right) \\) has area less than or equal to \\( mnbvcxza \\). Thus \\( \\frac{1}{2} asdfghjk\\, lkjhgfds \\leqq zxcvbnml \\) or\n\\[\nlkjhgfds \\leqq \\frac{2 zxcvbnml}{asdfghjk} .\n\\]\n\nEquality is possible and gives the maximum value of \\( lkjhgfds \\) (for given \\( zxcvbnml \\) and \\( asdfghjk \\) ) when the graph of \\( qzxwvtnp(mbvcxzpo) \\) is the straight line \\( qzxwvtnp(mbvcxzpo)=\\left(asdfghjk / lkjhgfds\\right) mbvcxzpo=\\left(asdfghjk^{2} / 2 zxcvbnml\\right) mbvcxzpo \\).\n\nComment: If \\( asdfghjk \\) is zero or negative, there is no maximum time \\( lkjhgfds \\) for the traverse. In the case \\( asdfghjk=0 \\) the equation of motion\n\\[\npoiuytre=zxcvbnml\\left[3\\left(mbvcxzpo / lkjhgfds\\right)^{2}-3\\left(mbvcxzpo / lkjhgfds\\right)^{3}\\right], \\quad 0 \\leqq mbvcxzpo \\leqq lkjhgfds\n\\]\nsatisfies the conditions of the problem for any \\( lkjhgfds>0 \\)."
    },
    "kernel_variant": {
      "question": "Let $m\\ge 2$.  \nFor every coordinate $k=1,\\dots ,m$ six positive real numbers  \n\n\\[\nu_k,\\;A_k,\\;B_k,\\;V_k,\\;D_k ,\\qquad 0<B_k<A_k ,\n\\]\n\nare prescribed.  \nThey are called, respectively, initial velocity, upper and lower acceleration\nbounds, required velocity gain and required extra displacement.\n\nFor $t\\ge 0$ we call an $m$–tuple  \n$v=(v_1,\\dots ,v_m)$ of piece–wise $C^{1}$ functions an admissible {\\bf\nvelocity profile} if for every coordinate $k$\n\n\\[\n\\dot x_k(t)=v_k(t),\\qquad a_k(t)=\\dot v_k(t)\n\\]\n\nsatisfies  \n\n(R1) $a_k:[0,\\infty)\\to\\mathbb R$ is bounded, measurable and\nnon–increasing;  \n\n(R2) $A_k\\ge a_k(t)\\ge B_k$ for all $t\\ge 0$ and $a_k(0)=A_k$  \n(the value at $0$ may differ from $a_k(0+)$, so downward jumps at $t=0$ are\nallowed).\n\nThe profile meets the prescribed terminal conditions at the instant $T>0$ when  \n\n\\[\nv_k(T)=u_k+V_k,\\qquad x_k(T)=u_k T+D_k\\qquad (k=1,\\dots ,m). \\tag{$\\star$}\n\\]\n\n1.  Give, solely in terms of the data $\\{A_k,B_k,V_k,D_k\\}$,  \n\n   (a) necessary and sufficient conditions for the existence of at least one\n       admissible profile verifying $(\\star)$ at some time $T>0$;  \n\n   (b) the complete set of times $T$ for which such a profile exists\n       simultaneously in every coordinate.\n\n2.  Whenever an admissible profile exists in all coordinates, determine  \n\n\\[\nT_{\\max}= \\max\\{T>0\\;:\\; (\\star)\\text{ can hold for every }k\\},\n\\]\n\nprove that this extremal time is unique, and construct an admissible profile\nwhich attains it.\n\n3.  Prove that no admissible profile can satisfy $(\\star)$ at any time\n$T>T_{\\max}$.",
      "solution": "Throughout we analyse one coordinate, suppress the index $k$ and finally\nsynchronise the $m$ coordinates.\n\nNotation for one coordinate  \n\n\\[\nu,\\;A,\\;B,\\;V,\\;D,\\qquad 0<B<A,\\qquad\n\\Delta=A-B,\\qquad v(0)=u .\n\\]\n\nBy translating the position variable we may assume $u=0$.\n\nI.  One–coordinate analysis  \n-------------------------------------------------\n\nI.1  Integral identities  \nFor every admissible acceleration $a(t)$\n\n\\[\nv(t)=\\int_{0}^{t} a(s)\\,ds,\\qquad \nx(t)=\\int_{0}^{t}(t-s)\\,a(s)\\,ds. \\tag{1}\n\\]\n\nConsequently the terminal requirements at some $T>0$ are\n\n\\[\n\\int_{0}^{T} a(s)\\,ds=V,\\qquad \n\\int_{0}^{T}(T-s)a(s)\\,ds=D. \\tag{2}\n\\]\n\nI.2  Two extremal envelopes  \n\nLower envelope.\nRearrangement (Chebyshev) gives for every admissible $a$\n\n\\[\nx(T)=\\int_{0}^{T}(T-s)a(s)\\,ds\n\\;\\ge\\;\\frac{1}{T}\\Bigl(\\int_{0}^{T}(T-s)\\,ds\\Bigr)\n        \\Bigl(\\int_{0}^{T} a(s)\\,ds\\Bigr)\n      =\\frac{V\\,T}{2}=:\\,L(T). \\tag{3}\n\\]\n\nThe value $L(T)$ is an {\\em infimum}.  \nIt is attained exactly when $a(0)=A$ and $a(t)\\equiv V/T$ for $t>0$.\nThis function is admissible because it is non–increasing,\n$B\\le V/T\\le A$ (see (5) below) and satisfies the prescribed jump at $t=0$.\n\nUpper envelope.  \nBecause the weight $T-s$ is strictly decreasing,  \n$\\Phi_T(a):=\\int_{0}^{T}(T-s)a(s)\\,ds$ is maximised, for fixed $V$,\nby moving as much acceleration as early as possible.  \nThe optimal bang–bang profile is\n\n\\[\na(t)=\\begin{cases}\nA,& 0<t<\\tau,\\\\\nB,& \\tau<t<T,\n\\end{cases}\n\\qquad \n\\tau=\\dfrac{V-BT}{\\Delta}. \\tag{4}\n\\]\n\nAdmissibility of (4) requires $0\\le \\tau\\le T$, namely\n\n\\[\n\\frac{V}{A}\\le T\\le\\frac{V}{B}. \\tag{5}\n\\]\n\nWith (4) one obtains\n\n\\[\n\\begin{aligned}\nU(T)&:=\\max_a x(T)=\nA\\int_{0}^{\\tau}(T-s)\\,ds+\nB\\int_{\\tau}^{T}(T-s)\\,ds\\\\\n&=V\\,T-\\frac{B\\,T^{2}}{2}-\\frac{(V-BT)^{2}}{2\\Delta},\\qquad\n\\frac{V}{A}\\le T\\le \\frac{V}{B}. \\tag{6}\n\\end{aligned}\n\\]\n\n$U$ is strictly increasing and strictly concave on $(V/A,V/B)$; in particular  \n\n\\[\nU\\!\\Bigl(\\tfrac{V}{A}\\Bigr)=\\frac{V^{2}}{2A},\\qquad \nU\\!\\Bigl(\\tfrac{V}{B}\\Bigr)=\\frac{V^{2}}{2B}. \\tag{7}\n\\]\n\nI.3  Feasible times and existence criterion  \n\nCombining (3), (5) and (6) we have\n\n\\[\nT\\text{ feasible }\\Longleftrightarrow\n\\frac{V}{A}\\le T\\le\\frac{V}{B}\\;\\hbox{ and }\\;\nL(T)\\le D\\le U(T). \\tag{8}\n\\]\n\nBecause $L$ is linear and $U$ is concave, the set\n\n\\[\nS=\\{T>0\\;:\\;L(T)\\le D\\le U(T)\\}\n\\]\n\nis an interval $[T_{*},T_{\\max}]$ or empty.\nWriting $L(T)\\le D$ gives $T\\le 2D/V$, hence\n\n\\[\nT_{\\max}=\\frac{2D}{V}. \\tag{9}\n\\]\n\nThe lower endpoint $T_{*}$ is the smaller root of $U(T)=D$ inside the band\n$[V/A,V/B]$:\n\n\\[\nA B\\,T^{2}-2A V\\,T+V^{2}+2\\Delta D=0. \\tag{10}\n\\]\n\nIts discriminant equals  \n$\\Delta_T=4A\\Delta\\,(V^{2}-2B D)$, therefore a real root exists iff\n\n\\[\n2B D\\le V^{2}\\le 2A D. \\tag{11}\n\\]\n\nWhen (11) holds the smaller root is\n\n\\[\nT_{*}=\\frac{A V-\\sqrt{A\\Delta\\,(V^{2}-2B D)}}{A B}. \\tag{12}\n\\]\n\nProposition 1 (single coordinate).  \nAn admissible profile exists iff (11) holds.  \nIf so, every feasible time lies in  \n\n\\[\nI=[T_{*},T_{\\max}]\n=\\Bigl[\\tfrac{A V-\\sqrt{A\\Delta\\,(V^{2}-2B D)}}{A B},\\;\n       \\tfrac{2D}{V}\\Bigr]. \\tag{13}\n\\]\n\nI.4  Construction for an arbitrary feasible time  \n\nFix $T\\in I$.\nDefine  \n\n\\[\n\\tau_{\\max}=\\frac{V-BT}{\\Delta}\\quad(\\ge 0).\n\\]\n\nFor any $\\tau\\in[0,\\tau_{\\max}]$ put  \n\n\\[\nc(\\tau)=\\frac{V-A\\tau}{T-\\tau}\\qquad (B\\le c(\\tau)\\le A),\n\\]\n\nand choose the two–level acceleration\n\n\\[\na(0)=A,\\qquad\na(t)=\\begin{cases}\nA,&0<t<\\tau,\\\\\nc(\\tau),&\\tau\\le t\\le T.\n\\end{cases} \\tag{14}\n\\]\n\nThe integral constraint gives $\\int_{0}^{T}a=V$ by construction.\nA direct calculation shows\n\n\\[\nx(T;\\tau)=A\\Bigl(T\\tau-\\frac{\\tau^{2}}{2}\\Bigr)\n          +c(\\tau)\\,\\frac{(T-\\tau)^{2}}{2}. \\tag{15}\n\\]\n\nThe map $\\tau\\mapsto x(T;\\tau)$  \nis continuous, $x(T;0)=L(T)$ and $x(T;\\tau_{\\max})=U(T)$.\nHence for every $D\\in[L(T),U(T)]$\nthere exists a (unique) $\\tau\\in[0,\\tau_{\\max}]$ with $x(T;\\tau)=D$.\nThus an admissible profile exists for every feasible $T$,\ncompleting the proof that $I$ is indeed the full feasible–time set.\n\nII.  Synchronising the $m$ coordinates  \n-------------------------------------------------\n\nFor every $k$ put  \n\n\\[\n\\Delta_k=A_k-B_k,\\qquad\nT^{(k)}_{*},T^{(k)}_{\\max}\\text{ as in (12),(9)}. \\tag{16}\n\\]\n\nCoordinate $k$ is feasible iff\n\n\\[\n2B_k D_k\\le V_k^{2}\\le 2A_k D_k\\quad\\text{(condition P$1_k$)}, \\tag{17}\n\\]\nand then it admits every time in the interval $I_k=[T^{(k)}_{*},T^{(k)}_{\\max}]$.\n\nThe vector motion is feasible at time $T$ iff $T\\in\\bigcap_{k=1}^{m}I_k$.\nConsequently\n\n\\[\nT_{\\max}:=\\min_{1\\le k\\le m}T^{(k)}_{\\max} \\tag{18}\n\\]\n\nis the largest {\\em candidate} time, and simultaneous feasibility is equivalent\nto  \n\n\\[\n\\text{(P$1_k$) for all }k,\\qquad \n\\max_{1\\le k\\le m}T^{(k)}_{*}\\;\\le\\;\n\\min_{1\\le k\\le m}T^{(k)}_{\\max}. \\tag{19}\n\\]\n\nIII.  Construction of an admissible profile at $T_{\\max}$  \n-------------------------------------------------\n\nAssume (19) holds and fix $T_{\\max}$ given by (18).\nFor each coordinate $k$ choose\n\n\\[\n\\tau_k\\in[0,\\;\\tau_{\\max,k}],\\qquad\n\\tau_{\\max,k}=\\frac{V_k-B_k T_{\\max}}{\\Delta_k},\n\\]\n\nsuch that $x_k(T_{\\max};\\tau_k)=D_k$ where $x_k(\\cdot;\\tau)$ is the map in\n(15).  \nThe existence and uniqueness of $\\tau_k$ follow from the discussion in I.4.\nDefine  \n\n\\[\na_k(0)=A_k,\\qquad\na_k(t)=\\begin{cases}\nA_k,&0<t<\\tau_k,\\\\\nc_k:=\\dfrac{V_k-A_k\\tau_k}{T_{\\max}-\\tau_k},&\\tau_k\\le t\\le T_{\\max},\\\\\nB_k,&t>T_{\\max}.\n\\end{cases}\n\\]\n\nBecause $B_k\\le c_k\\le A_k$, the function $a_k$ is admissible,\n$v_k(T_{\\max})=V_k$ and $x_k(T_{\\max})=D_k$.  \nThus the $m$–tuple $(a_1,\\dots ,a_m)$ furnishes an admissible\nvector profile achieving all terminal conditions at the common\ntime $T_{\\max}$.\n\nIV.  Optimality of $T_{\\max}$  \n-------------------------------------------------\n\nLet $T>T_{\\max}$ and choose an index $k_0$ with\n$T_{\\max}=T^{(k_0)}_{\\max}=2D_{k_0}/V_{k_0}$  \n(the minimum in (18) is attained).\nThen\n\n\\[\nL_{k_0}(T)=\\frac{V_{k_0}T}{2}\\;>\\;\n\\frac{V_{k_0}T^{(k_0)}_{\\max}}{2}=D_{k_0}. \\tag{20}\n\\]\n\nBy (3) the displacement $D_{k_0}$ cannot be produced at the later\ntime $T$, so no admissible vector profile can satisfy $(\\star)$.\nTherefore the value (18) is indeed the unique largest feasible time.\n\nThis completes the solution of items 1–3.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.606083",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher structural complexity  \n   • The motion is now in ℝᵐ with coordinate-wise but different upper and lower acceleration bounds (A_k and B_k).  \n   • The accelerations are required to be monotone *and* locked between two distinct positive levels, introducing interacting upper and lower constraints.  \n   • Both the first and the second integral moments of a(t) are prescribed (velocity and displacement), producing a genuine two-moment optimisation problem.\n\n2. Deeper theory  \n   • The solution needs a functional-analytic extreme-point argument (the bang–bang principle) on an infinite-dimensional convex set of admissible accelerations.  \n   • A rearrangement / Chebyshev inequality is invoked to justify that extremals concentrate the larger acceleration values near t=0.  \n   • The resulting optimisation reduces to manipulating coupled nonlinear equations, leading to a quadratic whose discriminant itself contains the data in a non-trivial way.\n\n3. Additional constraints and interacting concepts  \n   • The presence of both an upper and a lower bound on acceleration forces the extremal profile to have *two* distinct constant levels rather than the single-level solution of the original problem.  \n   • Feasibility now demands the non-obvious condition 2A_k D_k ≥ V_k²; recognising and proving this necessity is an extra step absent from the original.  \n   • Finally, the global answer is the *minimum* of m individual maxima, coupling m one-dimensional optimisation subproblems into a single multi-constraint extremal problem.\n\nAll these layers make the enhanced problem substantially harder: it demands variational thinking, functional inequalities, piece-wise optimisation, and the handling of non-trivial algebraic conditions—well beyond the elementary area‐under-curve argument of the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let $m\\ge 2$.  For every coordinate $k=1,\\dots ,m$ six positive parameters are prescribed  \n\n\\[\nu_k,\\;A_k,\\;B_k,\\;V_k,\\;D_k\\qquad (0<B_k<A_k),\n\\]\n\ncalled, respectively, initial velocity, initial upper acceleration bound, lower acceleration bound, required velocity gain, and required extra displacement.  \n\nFor $t\\ge 0$ we call an $m$–tuple $v=(v_1,\\dots ,v_m)$ of piece–wise $C^{1}$ (hence absolutely continuous) functions an admissible velocity profile if, for every coordinate $k$,\n\n\\[\n\\dot x_k(t)=v_k(t),\\qquad a_k(t)=\\dot v_k(t),\n\\]\n\nsatisfies  \n\n(R1) each $a_k$ is bounded, measurable and non–increasing on $[0,\\infty)$;  \n\n(R2) $A_k\\ge a_k(t)\\ge B_k$ for all $t\\ge 0$ and $a_k(0)=A_k$.  \n(The value $a_k(0)$ may differ from $a_k(0+)$, so instantaneous downward jumps are allowed.)\n\nThe profile meets the prescribed terminal conditions at the instant $T>0$ when  \n\n\\[\nv_k(T)=u_k+V_k,\\qquad x_k(T)=u_k\\,T+D_k\\qquad(k=1,\\dots ,m). \\tag{$\\star$}\n\\]\n\n1.  Express, solely in terms of the data $\\{A_k,B_k,V_k,D_k\\}$,  \n\n   (a) the necessary and sufficient conditions for the existence of at least one admissible profile verifying $(\\star)$ at some time $T>0$;  \n\n   (b) the complete set of times $T$ for which such a profile exists in every single coordinate.\n\n2.  When an admissible profile exists in all coordinates, determine  \n\n\\[\nT_{\\max}=\\hbox{the (unique) largest time for which }(\\star)\\hbox{ can hold},\n\\]\n\nand construct an admissible profile that realises $T_{\\max}$.\n\n3.  Prove that no admissible profile can satisfy $(\\star)$ at any time $T>T_{\\max}$.",
      "solution": "Throughout we write all formulas for a single coordinate and finally synchronise the $m$ coordinates.\n\nNotation for one coordinate.  \nFix $k$ and suppress the index:\n\n\\[\nu=u_k,\\;A=A_k,\\;B=B_k,\\;V=V_k,\\;D=D_k,\\;\\Delta=A-B>0,\\;v(0)=u.\n\\]\n\nBy translating the position variable we may assume $u=0$.\n\nI.  One–coordinate analysis  \n-------------------------------------------------\n\nI.1  Integral identities  \nFor every admissible acceleration $a(t)$\n\n\\[\nv(t)=\\int_{0}^{t} a(s)\\,ds,\\qquad \nx(t)=\\int_{0}^{t}(t-s)a(s)\\,ds. \\tag{1}\n\\]\n\nHence the terminal requirements at some $T>0$ read\n\n\\[\n\\int_{0}^{T} a(s)\\,ds=V,\\qquad \n\\int_{0}^{T}(T-s)a(s)\\,ds=D. \\tag{2}\n\\]\n\nDefine\n\n\\[\nL(T):=\\frac{V\\,T}{2}\\quad\\text{(minimal displacement at time $T$)},\\qquad \nU(T):=\\max_{a\\text{ admissible, (2) holds}} x(T).\\tag{3}\n\\]\n\nI.2  The upper envelope $U(T)$  \nBecause $(T-s)$ is strictly decreasing in $s$, the functional\n$\\Phi_T(a)=\\int_{0}^{T}(T-s)\\,a(s)\\,ds$ is maximised, for fixed $V$, by placing as much\nacceleration as possible near the origin.  \nThe optimal (bang–bang) profile is  \n\n\\[\na(t)=\\begin{cases}\nA,& 0\\le t\\le \\tau,\\\\[2pt]\nB,& \\tau\\le t\\le T,\n\\end{cases}\\qquad \n\\tau=\\dfrac{V-BT}{\\Delta}. \\tag{4}\n\\]\n\nAdmissibility of (4) requires $0\\le\\tau\\le T$, i.e.\n\n\\[\n\\frac{V}{A}\\le T\\le\\frac{V}{B}. \\tag{5}\n\\]\n\nWith (4) one obtains\n\n\\[\n\\begin{aligned}\nx(T)&=\\Delta T\\tau+\\frac{B\\,T^{2}}{2}-\\frac{\\Delta \\tau^{2}}{2} \\\\\n&=V\\,T-\\frac{B\\,T^{2}}{2}-\\frac{(V-BT)^{2}}{2\\Delta}.\n\\end{aligned}\n\\]\n\nThus  \n\n\\[\nU(T)=V\\,T-\\frac{B\\,T^{2}}{2}-\\frac{(V-BT)^{2}}{2\\Delta},\\qquad \\frac{V}{A}\\le T\\le\\frac{V}{B}. \\tag{6}\n\\]\n\n$U$ is strictly increasing and strictly concave on $(V/A,V/B)$; in particular  \n\n\\[\nU\\!\\Bigl(\\frac{V}{A}\\Bigr)=\\frac{V^{2}}{2A},\\qquad \nU\\!\\Bigl(\\frac{V}{B}\\Bigr)=\\frac{V^{2}}{2B}. \\tag{7}\n\\]\n\nI.3  The lower envelope $L(T)$  \nChebyshev’s inequality (or any rearrangement argument) applied to the decreasing\nsequence $T-s$ and the non–increasing $a(s)$ gives\n\n\\[\nx(T)=\\int_{0}^{T}(T-s)a(s)\\,ds\\;\\ge\\;\\frac{V\\,T}{2}=L(T), \\tag{8}\n\\]\n\nand the inequality is sharp (take $a(t)\\equiv V/T$).  \n\nI.4  Feasible times for one coordinate  \n\nCombining (5), (6) and (8) we have  \n\n\\[\nT\\text{ feasible}\\;\\Longleftrightarrow\\;\n\\frac{V}{A}\\le T\\le\\frac{V}{B},\\quad \nL(T)\\le D\\le U(T). \\tag{9}\n\\]\n\nBecause $L$ is linear increasing and $U$ is strictly concave increasing,\nthe feasible–time set is either empty or an interval $[T_{*},T_{\\max}]$.\n\nUpper end–point.  \nThe inequality $L(T)\\le D$ forces $T\\le 2D/V$, hence\n\n\\[\nT_{\\max}=\\frac{2D}{V}. \\tag{10}\n\\]\n\nLower end–point.  \n$T_{*}$ is the smaller root (inside the interval (5)) of $U(T)=D$:\n\n\\[\nA B\\,T^{2}-2A V\\,T+V^{2}+2\\Delta D=0. \\tag{11}\n\\]\n\nWriting the discriminant $\\Delta_T$ gives\n\n\\[\n\\Delta_T=4A\\Delta\\,(V^{2}-2B D),\n\\]\n\nso a real root exists iff  \n\n\\[\nV^{2}\\ge 2B D. \\tag{12}\n\\]\n\nThe smaller root is  \n\n\\[\nT_{*}=\\frac{A V-\\sqrt{A\\Delta\\,(V^{2}-2B D)}}{A B}. \\tag{13}\n\\]\n\nExistence criterion.  \nReality of (13) and inclusion $T_{*}\\ge V/A$ together are equivalent to  \n\n\\[\n2B D\\le V^{2}\\le 2A D. \\tag{14}\n\\]\n\nTherefore:\n\nProposition (one coordinate).  \nAn admissible profile exists iff (14) holds.  \nIf so, every feasible time lies in the interval  \n\n\\[\nI=[T_{*},T_{\\max}]\n=\\Bigl[\\frac{A V-\\sqrt{A\\Delta\\,(V^{2}-2B D)}}{A B},\\;\n\\frac{2D}{V}\\Bigr]. \\tag{15}\n\\]\n\nII.  Synchronising the $m$ coordinates  \n-------------------------------------------------\n\nFor every $k$ put  \n\n\\[\n\\Delta_k=A_k-B_k,\\quad \nT^{(k)}_{*}=\\frac{A_k V_k-\\sqrt{A_k\\Delta_k\\,(V_k^{2}-2B_k D_k)}}{A_k B_k},\\quad\nT^{(k)}_{\\max}=\\frac{2D_k}{V_k}. \\tag{16}\n\\]\n\nCoordinate $k$ is feasible iff  \n\n\\[\n2B_k D_k\\le V_k^{2}\\le 2A_k D_k, \\tag{P1$_k$}\n\\]\nand then it admits every time in $I_k=[T^{(k)}_{*},T^{(k)}_{\\max}]$.\n\nThe vector motion is feasible at time $T$ iff $T\\in\\bigcap_{k=1}^{m}I_k$.  \nConsequently\n\n\\[\nT_{\\max}:=\\min_{1\\le k\\le m}T^{(k)}_{\\max} \\tag{17}\n\\]\n\nis the largest candidate time, and simultaneous feasibility is equivalent to  \n\n\\[\n\\text{(P1$_k$) for every }k,\\qquad \n\\max_{1\\le k\\le m}T^{(k)}_{*}\\le\\min_{1\\le k\\le m}T^{(k)}_{\\max}. \\tag{18}\n\\]\n\nIII.  Construction of an admissible profile at $T_{\\max}$  \n-------------------------------------------------\n\nFix $T_{\\max}$ given by (17) and assume (18) holds.  \nFor each coordinate $k$ exactly one of the following mutually exclusive\nsituations occurs.\n\nCase A: $T_{\\max}<T^{(k)}_{\\max}$.  \nThen $T_{\\max}\\ge T^{(k)}_{*}$ and (4) gives a non–negative switch time  \n\n\\[\n\\tau_k=\\frac{V_k-B_k T_{\\max}}{A_k-B_k}\\in[0,T_{\\max}].\n\\]\n\nChoose the bang–bang profile (4) with this $\\tau_k$; it yields\n$v_k(T_{\\max})=V_k,\\;x_k(T_{\\max})=D_k$.\n\nCase B: $T_{\\max}=T^{(k)}_{\\max}=2D_k/V_k$ and $V_k/B_k>T_{\\max}$.  \nThen the constant value $a_k=V_k/T_{\\max}$ satisfies $B_k\\le a_k\\le A_k$.  \nSet  \n\n\\[\na_k(0)=A_k,\\qquad a_k(t)=\\frac{V_k}{T_{\\max}}\\quad(0<t\\le T_{\\max}). \\tag{19}\n\\]\n\nCase C (limit case): $T_{\\max}=V_k/B_k=2D_k/V_k$ (equivalently $V_k^{2}=2B_k D_k$).  \nTake $a_k(t)\\equiv B_k\\;(t>0)$.  \nAgain $v_k(T_{\\max})=V_k,\\;x_k(T_{\\max})=D_k$.\n\nCombining the $m$ coordinates furnishes an admissible vector\nprofile satisfying all terminal conditions at the common time $T_{\\max}$.\n\nIV.  Optimality of $T_{\\max}$  \n-------------------------------------------------\n\nLet $T>T_{\\max}$ and choose a coordinate $k_{0}$ with\n$T_{\\max}=T^{(k_0)}_{\\max}=2D_{k_0}/V_{k_0}$.\nThen $T>T^{(k_0)}_{\\max}$, so by (8)\n\n\\[\nL_{k_0}(T)=\\frac{V_{k_0}T}{2}>\\frac{V_{k_0}T^{(k_0)}_{\\max}}{2}=D_{k_0}. \\tag{20}\n\\]\n\nConsequently, for coordinate $k_{0}$ no admissible acceleration can produce the\nrequired displacement $D_{k_0}$ at the later time $T$.  Hence no admissible\nvector profile can satisfy $(\\star)$ for any $T>T_{\\max}$; the value\n(17) is therefore maximal.\n\nThis completes the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.484814",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher structural complexity  \n   • The motion is now in ℝᵐ with coordinate-wise but different upper and lower acceleration bounds (A_k and B_k).  \n   • The accelerations are required to be monotone *and* locked between two distinct positive levels, introducing interacting upper and lower constraints.  \n   • Both the first and the second integral moments of a(t) are prescribed (velocity and displacement), producing a genuine two-moment optimisation problem.\n\n2. Deeper theory  \n   • The solution needs a functional-analytic extreme-point argument (the bang–bang principle) on an infinite-dimensional convex set of admissible accelerations.  \n   • A rearrangement / Chebyshev inequality is invoked to justify that extremals concentrate the larger acceleration values near t=0.  \n   • The resulting optimisation reduces to manipulating coupled nonlinear equations, leading to a quadratic whose discriminant itself contains the data in a non-trivial way.\n\n3. Additional constraints and interacting concepts  \n   • The presence of both an upper and a lower bound on acceleration forces the extremal profile to have *two* distinct constant levels rather than the single-level solution of the original problem.  \n   • Feasibility now demands the non-obvious condition 2A_k D_k ≥ V_k²; recognising and proving this necessity is an extra step absent from the original.  \n   • Finally, the global answer is the *minimum* of m individual maxima, coupling m one-dimensional optimisation subproblems into a single multi-constraint extremal problem.\n\nAll these layers make the enhanced problem substantially harder: it demands variational thinking, functional inequalities, piece-wise optimisation, and the handling of non-trivial algebraic conditions—well beyond the elementary area‐under-curve argument of the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}