Answer
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Hint: Try to find what is asked in each part by doing proper solving for each and every part separately as per the requirement. Find the average acceleration for part (a), position of the train for part (c) and for part (b) and (d), try to find what is asked accordingly.
Complete step by step answer:
a) Average acceleration
$\begin{align}
& =\dfrac{{{\left[ {{v}_{A}}\left( t \right) \right]}_{f}}-{{\left[ {{v}_{A}}\left( t \right) \right]}_{i}}}{{{\left[ t \right]}_{f}}-{{\left[ t \right]}_{i}}} \\
& =\dfrac{{{v}_{A}}\left( 8 \right)-{{v}_{A}}\left( 2 \right)}{8-2} \\
& =\dfrac{-120-100}{6} \\
& =-\dfrac{110}{3}meter/{{\min }^{2}} \\
\end{align}$
b) Yes, as the function ${{v}_{A}}\left( t \right)$is differentiable and thus a continuous function, so the velocity of train A must at some time ‘t’ within 5 < t < 8 equal to -100 meters/min because ${{v}_{A}}$(t)=-100 lies between ${{v}_{A}}$(5)=40 and ${{v}_{A}}$(8)=-120
c)
\[\begin{align}
& x\left( 12 \right)=\int\limits_{2}^{12}{{{v}_{A}}\left( t \right)dt}+x\left( 2 \right) \\
& \Rightarrow x\left( 12 \right)=3.\dfrac{1}{2}.\left( 100+40 \right)+3.\dfrac{1}{2}.\left( 40-120 \right)+4.\dfrac{1}{2}.\left( -120-150 \right)+300 \\
& \Rightarrow x\left( 12 \right)=210-120-540+300 \\
& \Rightarrow x\left( 12 \right)=-150 \\
\end{align}\]
So, at t= 12 minute the train is approximately 150 meters west of origin station
d) Let x be train A’s position, y train B’s position, and z the distance between train A and train B.
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}={{z}^{2}} \\
& \\
\end{align}\]
Differentiating wrt ‘t’;
$\begin{align}
& \Rightarrow 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=2z\dfrac{dz}{dt} \\
& Here;x=300,\text{ }y=400 \\
& z=\sqrt{{{300}^{2}}+{{400}^{2}}}=500 \\
& {{v}_{B}}\left( 2 \right)=-20+120+25=125 \\
& \Rightarrow 300\times 100+400\times 125=500\times \dfrac{dz}{dt} \\
& \Rightarrow \dfrac{dz}{dt}=\dfrac{80000}{500}=160meters/\min \\
\end{align}$
Note: Calculation should be done according to the requirement of each part of the question. As in part (a) we have to find the average acceleration, in part (c) the position of the train and in part (d) the differentiation.
Complete step by step answer:
a) Average acceleration
$\begin{align}
& =\dfrac{{{\left[ {{v}_{A}}\left( t \right) \right]}_{f}}-{{\left[ {{v}_{A}}\left( t \right) \right]}_{i}}}{{{\left[ t \right]}_{f}}-{{\left[ t \right]}_{i}}} \\
& =\dfrac{{{v}_{A}}\left( 8 \right)-{{v}_{A}}\left( 2 \right)}{8-2} \\
& =\dfrac{-120-100}{6} \\
& =-\dfrac{110}{3}meter/{{\min }^{2}} \\
\end{align}$
b) Yes, as the function ${{v}_{A}}\left( t \right)$is differentiable and thus a continuous function, so the velocity of train A must at some time ‘t’ within 5 < t < 8 equal to -100 meters/min because ${{v}_{A}}$(t)=-100 lies between ${{v}_{A}}$(5)=40 and ${{v}_{A}}$(8)=-120
c)
\[\begin{align}
& x\left( 12 \right)=\int\limits_{2}^{12}{{{v}_{A}}\left( t \right)dt}+x\left( 2 \right) \\
& \Rightarrow x\left( 12 \right)=3.\dfrac{1}{2}.\left( 100+40 \right)+3.\dfrac{1}{2}.\left( 40-120 \right)+4.\dfrac{1}{2}.\left( -120-150 \right)+300 \\
& \Rightarrow x\left( 12 \right)=210-120-540+300 \\
& \Rightarrow x\left( 12 \right)=-150 \\
\end{align}\]
So, at t= 12 minute the train is approximately 150 meters west of origin station
d) Let x be train A’s position, y train B’s position, and z the distance between train A and train B.
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}={{z}^{2}} \\
& \\
\end{align}\]
Differentiating wrt ‘t’;
$\begin{align}
& \Rightarrow 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=2z\dfrac{dz}{dt} \\
& Here;x=300,\text{ }y=400 \\
& z=\sqrt{{{300}^{2}}+{{400}^{2}}}=500 \\
& {{v}_{B}}\left( 2 \right)=-20+120+25=125 \\
& \Rightarrow 300\times 100+400\times 125=500\times \dfrac{dz}{dt} \\
& \Rightarrow \dfrac{dz}{dt}=\dfrac{80000}{500}=160meters/\min \\
\end{align}$
Note: Calculation should be done according to the requirement of each part of the question. As in part (a) we have to find the average acceleration, in part (c) the position of the train and in part (d) the differentiation.
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