QUESTION 1
Since we know exactly the dimensions of some expressions that are included in the formula, we can conclude
{ [ v ( t ) ] = [ m s e c ] [ 1 − e − t / t m a x s p e e d ] = [ just a number ] → [ A ] = [ m s e c ] \left\{\begin{array}{l}
[v(t)]=\left[\displaystyle\frac{m}{sec}\right]\\[0.3cm]
\left[1-e^{-t/t_{maxspeed}}\right]=[\text{just a number}]
\end{array}\right.\rightarrow [A]=\left[\frac{m}{sec}\right] ⎩ ⎨ ⎧ [ v ( t )] = [ sec m ] [ 1 − e − t / t ma x s p ee d ] = [ just a number ] → [ A ] = [ sec m ]
We substitute t = 0 t=0 t = 0 and find the value of velocity :
v ( 0 ) = A ⋅ ( 1 − e − 0 t m a x s p e e d ) = A ⋅ ( 1 − 1 ) = 0 v ( 0 ) = 0 v(0)=A\cdot\left(1-e^{-\displaystyle\frac{0}{t_{maxspeed}}}\right)=A\cdot(1-1)=0\\[0.3cm]
\boxed{v(0)=0} v ( 0 ) = A ⋅ ⎝ ⎛ 1 − e − t ma x s p ee d 0 ⎠ ⎞ = A ⋅ ( 1 − 1 ) = 0 v ( 0 ) = 0
To find the asymptote as t → + ∞ t\to+\infty t → + ∞ we calculate the limit :
v ∞ = lim t → + ∞ ( A ⋅ ( 1 − e − t t m a x s p e e d ) ) = A ⋅ ( 1 − 0 ) = A v ∞ = A 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]
\boxed{v_\infty=A} v ∞ = t → + ∞ lim ⎝ ⎛ A ⋅ ⎝ ⎛ 1 − e − t ma x s p ee d t ⎠ ⎞ ⎠ ⎞ = A ⋅ ( 1 − 0 ) = A v ∞ = A
Now we can conclude about the physical meaning of the constant A A A : A A A is the boundary speed for a given type of motion.
ANSWER
{ [ A ] = [ m s e c ] A − the boundary speed v ( 0 ) = 0 v ∞ = A \left\{\begin{array}{l}
[A]=\left[\displaystyle\frac{m}{sec}\right]\\[0.3cm]
A-\text{the boundary speed}\\[0.3cm]
v(0)=0\\[0.3cm]
v_\infty=A
\end{array}\right. ⎩ ⎨ ⎧ [ A ] = [ sec m ] A − the boundary speed v ( 0 ) = 0 v ∞ = A
QUESTION 2
To plot the graph, I chose the following constants:
A = 10 t m a x s p e e d = 5 A=10\\[0.3cm]
t_{maxspeed}=5 A = 10 t ma x s p ee d = 5
QUESTION 3
As we know
v ( t ) = d x ( t ) d t → x ( t ) = ∫ v ( t ) d x = ∫ ( A ⋅ ( 1 − e − t t m a x s p e e d ) ) d t x ( t ) = A ⋅ ( t + t m a x s p e e d ⋅ e − t t m a x s p e e d ) + C o n s t x ( t ) = A ⋅ t m a x s p e e d ⋅ ( t t m a x s p e e d + e − t t m a x s p e e d ) + C o n s t 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]
x(t)=A\cdot\left(t+t_{maxspeed}\cdot e^{-\displaystyle\frac{t}{t_{maxspeed}}}\right)+Const\\[0.3cm]
\boxed{x(t)=A\cdot t_{maxspeed}\cdot\left(\frac{t}{t_{maxspeed}}+e^{-\displaystyle\frac{t}{t_{maxspeed}}}\right)+Const} v ( t ) = d t d x ( t ) → x ( t ) = ∫ v ( t ) d x = ∫ ⎝ ⎛ A ⋅ ⎝ ⎛ 1 − e − t ma x s p ee d t ⎠ ⎞ ⎠ ⎞ d t x ( t ) = A ⋅ ⎝ ⎛ t + t ma x s p ee d ⋅ e − t ma x s p ee d t ⎠ ⎞ + C o n s t x ( t ) = A ⋅ t ma x s p ee d ⋅ ⎝ ⎛ t ma x s p ee d t + e − t ma x s p ee d t ⎠ ⎞ + C o n s t
To find the starting position, we will assume that C o n s t = 0 Const=0 C o n s t = 0 , then
x ( 0 ) = A ⋅ t m a x s p e e d ⋅ ( 0 t m a x s p e e d + e − 0 t m a x s p e e d ) x ( 0 ) = A ⋅ t m a x s p e e d x(0)=A\cdot t_{maxspeed}\cdot\left(\frac{0}{t_{maxspeed}}+e^{-\displaystyle\frac{0}{t_{maxspeed}}}\right)\\[0.3cm]
\boxed{x(0)=A\cdot t_{maxspeed}} x ( 0 ) = A ⋅ t ma x s p ee d ⋅ ⎝ ⎛ t ma x s p ee d 0 + e − t ma x s p ee d 0 ⎠ ⎞ x ( 0 ) = A ⋅ t ma x s p ee d
Using the same assumption C o n s t = 0 Const=0 C o n s t = 0 , we can find the asymptote for t → + ∞ t\to+\infty t → + ∞ :
x ∞ = lim t → + ∞ ( A ⋅ t m a x s p e e d ⋅ ( t t m a x s p e e d + e − t t m a x s p e e d ) ) = + ∞ x ∞ = + ∞ 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]
\boxed{x_\infty=+\infty} x ∞ = t → + ∞ lim ⎝ ⎛ A ⋅ t ma x s p ee d ⋅ ⎝ ⎛ t ma x s p ee d t + e − t ma x s p ee d t ⎠ ⎞ ⎠ ⎞ = + ∞ x ∞ = + ∞
ANSWER
General equation of motion:
x ( t ) = A ⋅ t m a x s p e e d ⋅ ( t t m a x s p e e d + e − t t m a x s p e e d ) + C o n s t x(t)=A\cdot t_{maxspeed}\cdot\left(\frac{t}{t_{maxspeed}}+e^{-\displaystyle\frac{t}{t_{maxspeed}}}\right)+Const x ( t ) = A ⋅ t ma x s p ee d ⋅ ⎝ ⎛ t ma x s p ee d t + e − t ma x s p ee d t ⎠ ⎞ + C o n s t
Under the assumption that C o n s t = 0 Const=0 C o n s t = 0 :
{ x ( 0 ) = A ⋅ t m a x s p e e d x ∞ = + ∞ \left\{\begin{array}{l}
x(0)=A\cdot t_{maxspeed}\\[0.3cm]
x_\infty=+\infty
\end{array}\right. { x ( 0 ) = A ⋅ t ma x s p ee d x ∞ = + ∞
QUESTION 4
To plot the graph, I chose the following constants:
A = 10 t m a x s p e e d = 5 A=10\\[0.3cm]
t_{maxspeed}=5 A = 10 t ma x s p ee d = 5
QUESTION 5
As we know
a ( t ) = d v ( t ) d t = d d t ( A ⋅ ( 1 − e − t t m a x s p e e d ) ) a ( t ) = A t m a x s p e e d ⋅ e − t t m a x s p e e d a(t)=\frac{dv(t)}{dt}=\frac{d}{dt}\left(A\cdot\left(1-e^{-\displaystyle\frac{t}{t_{maxspeed}}}\right)\right)\\[0.3cm]
\boxed{a(t)=\frac{A}{t_{maxspeed}}\cdot e^{-\displaystyle\frac{t}{t_{maxspeed}}}} a ( t ) = d t d v ( t ) = d t d ⎝ ⎛ A ⋅ ⎝ ⎛ 1 − e − t ma x s p ee d t ⎠ ⎞ ⎠ ⎞ a ( t ) = t ma x s p ee d A ⋅ e − t ma x s p ee d t
Initial acceleration is
a ( 0 ) = A t m a x s p e e d ⋅ e − 0 t m a x s p e e d a ( 0 ) = A t m a x s p e e d a(0)=\frac{A}{t_{maxspeed}}\cdot e^{-\displaystyle\frac{0}{t_{maxspeed}}}\\[0.3cm]
\boxed{a(0)=\frac{A}{t_{maxspeed}}} a ( 0 ) = t ma x s p ee d A ⋅ e − t ma x s p ee d 0 a ( 0 ) = t ma x s p ee d A
To find the asymptote as t → + ∞ t\to+\infty t → + ∞ we calculate the limit :
a ∞ = lim t → + ∞ ( A t m a x s p e e d ⋅ e − t t m a x s p e e d ) = 0 a ∞ = 0 a_\infty=\lim\limits_{t\to+\infty}\left(\frac{A}{t_{maxspeed}}\cdot e^{-\displaystyle\frac{t}{t_{maxspeed}}}\right)=0\\[0.3cm]
\boxed{a_\infty=0} a ∞ = t → + ∞ lim ⎝ ⎛ t ma x s p ee d A ⋅ e − t ma x s p ee d t ⎠ ⎞ = 0 a ∞ = 0
ANSWER
{ a ( t ) = A t m a x s p e e d ⋅ e − t t m a x s p e e d a ( 0 ) = A t m a x s p e e d a ∞ = 0 \left\{\begin{array}{l}
a(t)=\displaystyle\frac{A}{t_{maxspeed}}\cdot e^{-\displaystyle\frac{t}{t_{maxspeed}}}\\[0.3cm]
a(0)=\displaystyle\frac{A}{t_{maxspeed}}\\[0.3cm]
a_\infty=0
\end{array}\right. ⎩ ⎨ ⎧ a ( t ) = t ma x s p ee d A ⋅ e − t ma x s p ee d t a ( 0 ) = t ma x s p ee d A a ∞ = 0
QUESTION 6
To plot the graph, I chose the following constants:
A = 10 t m a x s p e e d = 5 A=10\\[0.3cm]
t_{maxspeed}=5 A = 10 t ma x s p ee d = 5
QUESTION 7
Initial conditions x ( 0 ) = 400 x(0)=400 x ( 0 ) = 400 and v ( 0 ) = 28 v(0)=28 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 m a x s p e e d t_{maxspeed} t ma x s p ee d , although from the same paragraph 2 we can conclude that t m a x s p e e d = + ∞ t_{maxspeed}=+\infty t ma x s p ee d = + ∞ , which is clearly not suitable for real tasks