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Day: July 27, 2019

Posted on July 27, 2019July 25, 2022

201907271536 Solution to 1991-CE-PHY-II-4

T=\displaystyle{\frac{1}{f}}=1/5=0.2\,\mathrm{s} when f=5\,\mathrm{Hz}.


Setup.

Given that
\begin{aligned}[t] \textrm{Time }t & \quad & t_1 & \quad & t_2=t_1+T & \quad & t_3=t_1+2T & \quad & t_4=t_1+3T \\ \textrm{DIsplacement }s & \quad & s_1=0 & \quad & s_2=s_1+4=4 &\quad & s_3=s_2+8=12 & \quad & s_4=s_3+12=24 \\ \textrm{Velocity }v & \quad & u_1 & \quad & u_2 & \quad & u_3 & \quad & u_4  \\ \textrm{Acceleration }a & \quad & a\textrm{ (= Const.)} & \quad & a\textrm{ (= Const.)} & \quad & a\textrm{ (= Const.)}& \quad & a\textrm{ (= Const.)} \end{aligned}
from the figure.


Doing substitution for s, u, a, and t in

s=ut+\displaystyle{\frac{1}{2}}at^2,

I have some equations:

\begin{aligned} 4 &= s_2 = u_1T + \frac{1}{2}aT^2 \\ 12 &= s_3 = u_1(2T) + \frac{1}{2}a(2T)^2 \\ 24 &= s_4 = u_1(3T) + \frac{1}{2}a(3T)^2 \\ \end{aligned}

Let a be the subject,

\begin{aligned} a & = \frac{8}{T^2} - \frac{2u}{T} \\ a & = \frac{6}{T^2} - \frac{u}{T} \\ a & = \frac{16}{3T^2} - \frac{2u}{3T} \end{aligned}

As is known T=0.2,

\begin{aligned} a & = 200- 10u \\ a & = 150 - 5u\\ a & = \frac{400}{3} - \frac{10u}{3} \end{aligned}

Solving,

u=10\enspace (\mathrm{cm\, s^{-1}}) and a=100\enspace (\mathrm{cm}).

And the answer is D.


Roughwork.

Unnecessarily tedious,

\begin{aligned} a+10u & =200 \\ a + 5u & =150 \end{aligned} \quad \Rightarrow \quad \begin{bmatrix} 1 & 10 \\ 1 & 5 \end{bmatrix} \begin{bmatrix} a \\ u   \end{bmatrix} = \begin{bmatrix} 200 \\ 150  \end{bmatrix}\quad  \Rightarrow\quad \begin{bmatrix} a \\ u \end{bmatrix} = \begin{bmatrix} 1 & 10 \\ 1 & 5 \end{bmatrix}^{-1} \begin{bmatrix} 200 \\ 150  \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 0.2 & -0.2 \end{bmatrix} \begin{bmatrix} 200 \\ 150  \end{bmatrix} = \begin{bmatrix} (-1)(200) + (2)(150) \\ (0.2)(200) + (-0.2)(150)  \end{bmatrix} = \begin{bmatrix} 100 \\ 10 \end{bmatrix},

be simple.

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