QUESTION 1
Since we know exactly the dimensions of some expressions that are included in the formula, we can conclude
"\\left\\{\\begin{array}{l}\n[v(t)]=\\left[\\displaystyle\\frac{m}{sec}\\right]\\\\[0.3cm]\n\\left[1-e^{-t\/t_{maxspeed}}\\right]=[\\text{just a number}]\n\\end{array}\\right.\\rightarrow [A]=\\left[\\frac{m}{sec}\\right]"
We substitute "t=0" and find the value of velocity :
"v(0)=A\\cdot\\left(1-e^{-\\displaystyle\\frac{0}{t_{maxspeed}}}\\right)=A\\cdot(1-1)=0\\\\[0.3cm]\n\\boxed{v(0)=0}"
To find the asymptote as "t\\to+\\infty" we calculate the limit :
"v_{\\infty}=\\lim\\limits_{t\\to+\\infty}\\left(A\\cdot\\left(1-e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)\\right)=A\\cdot(1-0)=A\\\\[0.3cm]\n\\boxed{v_\\infty=A}"
Now we can conclude about the physical meaning of the constant "A" : "A" is the boundary speed for a given type of motion.
ANSWER
"\\left\\{\\begin{array}{l}\n[A]=\\left[\\displaystyle\\frac{m}{sec}\\right]\\\\[0.3cm]\nA-\\text{the boundary speed}\\\\[0.3cm]\nv(0)=0\\\\[0.3cm]\nv_\\infty=A\n\\end{array}\\right."
QUESTION 2
To plot the graph, I chose the following constants:
"A=10\\\\[0.3cm]\nt_{maxspeed}=5"
QUESTION 3
As we know
"v(t)=\\frac{dx(t)}{dt}\\to x(t)=\\int v(t)dx=\\int \\left(A\\cdot\\left(1-e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)\\right)dt\\\\[0.3cm]\nx(t)=A\\cdot\\left(t+t_{maxspeed}\\cdot e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)+Const\\\\[0.3cm]\n\\boxed{x(t)=A\\cdot t_{maxspeed}\\cdot\\left(\\frac{t}{t_{maxspeed}}+e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)+Const}"
To find the starting position, we will assume that "Const=0" , then
"x(0)=A\\cdot t_{maxspeed}\\cdot\\left(\\frac{0}{t_{maxspeed}}+e^{-\\displaystyle\\frac{0}{t_{maxspeed}}}\\right)\\\\[0.3cm]\n\\boxed{x(0)=A\\cdot t_{maxspeed}}"
Using the same assumption "Const=0" , we can find the asymptote for "t\\to+\\infty" :
"x_\\infty=\\lim\\limits_{t\\to+\\infty}\\left(A\\cdot t_{maxspeed}\\cdot\\left(\\frac{t}{t_{maxspeed}}+e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)\\right)=+\\infty\\\\[0.3cm]\n\\boxed{x_\\infty=+\\infty}"
ANSWER
General equation of motion:
"x(t)=A\\cdot t_{maxspeed}\\cdot\\left(\\frac{t}{t_{maxspeed}}+e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)+Const"
Under the assumption that "Const=0" :
"\\left\\{\\begin{array}{l}\nx(0)=A\\cdot t_{maxspeed}\\\\[0.3cm]\nx_\\infty=+\\infty\n\\end{array}\\right."
QUESTION 4
To plot the graph, I chose the following constants:
"A=10\\\\[0.3cm]\nt_{maxspeed}=5"
QUESTION 5
As we know
"a(t)=\\frac{dv(t)}{dt}=\\frac{d}{dt}\\left(A\\cdot\\left(1-e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)\\right)\\\\[0.3cm]\n\\boxed{a(t)=\\frac{A}{t_{maxspeed}}\\cdot e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}}"
Initial acceleration is
"a(0)=\\frac{A}{t_{maxspeed}}\\cdot e^{-\\displaystyle\\frac{0}{t_{maxspeed}}}\\\\[0.3cm]\n\\boxed{a(0)=\\frac{A}{t_{maxspeed}}}"
To find the asymptote as "t\\to+\\infty" we calculate the limit :
"a_\\infty=\\lim\\limits_{t\\to+\\infty}\\left(\\frac{A}{t_{maxspeed}}\\cdot e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\right)=0\\\\[0.3cm]\n\\boxed{a_\\infty=0}"
ANSWER
"\\left\\{\\begin{array}{l}\na(t)=\\displaystyle\\frac{A}{t_{maxspeed}}\\cdot e^{-\\displaystyle\\frac{t}{t_{maxspeed}}}\\\\[0.3cm]\na(0)=\\displaystyle\\frac{A}{t_{maxspeed}}\\\\[0.3cm]\na_\\infty=0\n\\end{array}\\right."
QUESTION 6
To plot the graph, I chose the following constants:
"A=10\\\\[0.3cm]\nt_{maxspeed}=5"
QUESTION 7
Initial conditions "x(0)=400" and "v(0)=28" .
These initial conditions cannot be, since in point 2 we theoretically proved that the initial speed must be 0. Therefore, I can not answer this question. Moreover, you need to specify the value "t_{maxspeed}" , although from the same paragraph 2 we can conclude that "t_{maxspeed}=+\\infty" , which is clearly not suitable for real tasks
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