Question

Train A runs back and forth on an east-west section of railroad track. Train A’s velocity, measured in meters per minute, is given by a differentiable function ${{v}_{A}}\left( t \right)$ where time t is measured in minutes. Selected values for ${{v}_{A}}\left( t \right)$ are given in the table below.

t (Minutes)	0	2	5	8	12
${{v}_{A}}\left( t \right)$(meters/minute)	0	100	40	-120	-150

a) Find the average acceleration of train A over the interval 2 ≤ t ≤ 8.
b) Do the data in the table support the conclusion that train A’s velocity is −100 meters per minute at some time t with 5 < t < 8 ? Give a reason for your answer.
c) At time t = 2, train A’s position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train A, in meters from the Origin Station, at time t = 12. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time t = 12.
d) A second train, train B, travels north from the Origin Station. At time t the velocity of train B is given by ${{v}_{b}}\left( t \right)=-5{{t}^{2}}+60t+25$ and at time t = 2 the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train A and train B is changing at time t = 2.

Answer 1

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.