October 20, 2022 – Physicspupil

October 20, 2022October 24, 2022

202210201704 Solution to 1971-CE-AMATH-II-XX

A particle is projected vertically from the ground so that its velocity (in $\mathrm{ft/sec}$ ) in the first $4$ seconds is given by

$V_1=128-32t\qquad (0\leqslant t\leqslant 4)$

and after the fourth second by

$V_2=3t^2-32t+80\qquad (4\leqslant t)$

where $t$ is the time (in seconds) after projection.

i. Show that $V_1=V_2$ when $t=4$ .
ii. Calculate the height of the particle when $t=4$ .
iii. Calculate the height of the particle when $t=5$ .
iv. Find the value of $t$ when the acceleration of the particle is $40\,\mathrm{ft/sec^2}$ .

Roughwork.

i.

When $t=4$ :

$\begin{aligned} V_1 & = 128-32(4) \\ & = 0 \\ V_2 & = 3(4)^2-32(4)+80 \\ & = 0 \\ V_1 & = V_2 \\ \end{aligned}$

ii.

$\begin{aligned} h(t\in [0,4]) & = \int_{0}^{4}V_1\,\mathrm{d}t \\ & = \int_{0}^{4}(128-32t)\,\mathrm{d}t \\ & = \bigg[ 128\bigg(\frac{t^{(0)+1}}{(0)+1}\bigg) - 32\bigg(\frac{t^{(1)+1}}{(1)+1}\bigg)\bigg]\bigg|_{0}^{4} \\ & = [128(4)-16(4)^2]-[128(0)-16(0)^2] \\ & = 256\,\mathrm{ft} \\ \end{aligned}$

iii.

$\begin{aligned} h(t\in [0,5]) & = h(t\in[0,4]) + h(t\in[4,5]) \\ & = 256\,\mathrm{ft} + \int_{4}^{5}V_2\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \int_{4}^{5}(3t^2-32t+80)\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \big[ t^3-16t^2+80t\big]\big|_{4}^{5} \\ & = 256\,\mathrm{ft} -3\,\mathrm{ft} \\ & = 253\,\mathrm{ft} \\ \end{aligned}$

iv.

$\begin{aligned} a(t\in [0,4]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_1=-32 \\ a(t\in [4,T]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_2=6t-32 \\ \end{aligned}$

$\begin{aligned} 6t-32 & = 40 \\ t & = 12\,\mathrm{s}\\ \end{aligned}$

October 20, 2022October 28, 2022

202210201648 Solution to 1972-CE-AMATH-II-XX

It is given that $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=4x-\frac{3}{2}x^2}$ and that $y=10$ when $x=2$ . Find $y$ in terms of $x$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int\bigg( 4x-\frac{3}{2}x^2\bigg)\,\mathrm{d}x \\ & = \frac{4x^{(1)+1}}{(1)+1} -\frac{3}{2}\bigg(\frac{x^{(2)+1}}{(2)+1}\bigg) + C\textrm{ for some constant} \\ & = 2x^2+\frac{1}{2}x^3 + C \\ \end{aligned}$

$\begin{aligned} 10 & = 2(2)^2+\frac{1}{2}(2)^3+C \\ C & = -2 \\ \end{aligned}$

In conclusion,

$y(x)=2x^2+\frac{1}{2}x^3-2$ .

October 20, 2022October 20, 2022

202210201553 Solution to 1977-CE-AMATH-I-XX

In the figure below, the complex numbers $z_0$ , $z_1$ , $z_2$ , $z_3$ , and $z_4$ are represented in the Argand diagram by the vertices of a regular pentagon with centre at the origin $O$ . If $z_0=2$ , write $z_2$ in polar form and calculate the value of $(z_2)^5$ .

Recall.

A complex number $z$ in Cartesian form $z=a+\mathrm{i}b$ can also be expressed in the polar form

$z=re^{\mathrm{i}\varphi}=r(\cos\varphi +\mathrm{i}\sin\varphi )$

where $r=\sqrt{a^2+b^2}$ is called the modulus $|z|$ , and $\varphi$ the argument $\textrm{arg}(z)$ , of $z$ .

Roughwork.

Observe that

$|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=2$ .

and

$\displaystyle{\textrm{arg}(z_2)=2\pi\cdot\frac{2}{5}=\frac{4\pi}{5}}$ .

Hence

$z_2=2e^{\mathrm{i}\frac{4\pi}{5}}$ .

The remaining are left the reader as an exercise.

October 20, 2022October 20, 2022

202210201121 Solution to 1977-CE-AMATH-I-XX

Prove, by mathematical induction, that $23^n-1$ is divisible by $11$ .

Roughwork.

Let proposition $P(n)$ be that

$P(n):\enspace 11|23^n-1$ .

First, note

$\begin{aligned} & \quad\enspace P(1):\enspace 11|23^{1}-1 \\ & \Leftrightarrow P(1):\enspace 11|23-1 \\ & \Leftrightarrow P(1):\enspace 11|22 \\ & \Leftrightarrow P(1)\textrm{ is true} \\ \end{aligned}$

that $P(n)$ is true for $n=1$ .

Next, assumed be to $P(n)$ true.

$\begin{aligned} &\quad\enspace P(n+1):\enspace 11|23^{(n+1)}-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^{n}\cdot 23-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^n-1+23^n\cdot 22\\ & \dots\textrm{ by assumption }11|23^n-1\textrm{ and for }11|22 \\ & \Leftrightarrow P(n+1)\textrm{ is true} \\ \end{aligned}$

Thus, $P(n)$ is true implies $P(n+1)$ true.

From $P(1)\textrm{ true}\,\vee\, P(n)\Longrightarrow P(n+1)\textrm{ true}$ follows $P(n)$ the proposition is true for all positive integers $n\geqslant 1$ . By the principle of mathematical induction proven is that $11$ divides $23^n-1$ .