Velocity Time Graphs (Linear Motion) - Worked Example

1. A train, which starts its motion from rest, moves along a straight linear road. It  takes a velocity of $u$ by moving with an acceleration of $f_1$ and then a velocity of $3u$ by moving with an acceleration of $f_2$. After moving with this velocity for a period of $\cfrac{u}{f_1}$, it takes a deceleration of $f_2$ until it comes back to the rest. Show that the total displacement of the train is $\cfrac{u^2[17f_1+7f_2]}{2f_1 f_2}$.
Answer




On this graph, OA line segment represent the $f_1$ acceleration and AB represent the $f_2$ acceleration. At BC train moves with a constant velocity and then at CD it decelerate from $f_2$, as described in the question.
From the given data,
$\tan\alpha = f_1$
$\tan\beta = f_2$
OG $= u \cot\alpha = \cfrac{u}{f_1}$
GF $= (3u-u)\cot\beta = \cfrac{2u}{f_2}$
ED $= 3u \cot\beta = \cfrac{3u}{f_2}$

The total displacement = OABCD area
$\therefore$ Total displacement = $\cfrac{1}{2}$OG$\cdot$GA + $\cfrac{1}{2}$(AG+BF)$\cdot$GF + BF$\cdot$EF + $\cfrac{1}{2}$CE$\cdot$ED
                                    $= \cfrac{1}{2}\cdot u \cdot \cfrac{u}{f_1}+\cfrac{3u+u}{2}\cdot \cfrac{2u}{f_2}+3u\cdot \cfrac{u}{f_1}+\cfrac{1}{2}\cdot 3u \cdot \cfrac{3u}{f_2}$
                                    $= \cfrac{u^2}{2f_1}+\cfrac{8u^2}{2f_2}+\cfrac{3u^2}{f_1}+\cfrac{9u^2}{2f_2}$
                                    $=\cfrac{7u^2}{2f_1}+\cfrac{17u^2}{2f_2}$
                                    $=\cfrac{u^2[17f_1+7f_2]}{2f_1 f_2}$


2. To be continued... 

Answers For Cambridge AS & A Level P1 (Pure) 2014 (9709/11)

*These answers are only for reference purposes and are not the official marking scheme for the relevant paper. Download Paper

1. $(2+ax)^7=\displaystyle\sum_{r=0}^{7} {}^7C_r 2^{7-r} (ax)^r$
   The terms of $x$ and $x^2$ can get by $r=1$ and $r=2$ respectively.
   $T_1={}^7C_1 2^{7-1} (ax)$ and $T_2={}^7C_2 2^{7-2} (ax)^2$
   $\therefore {}^7C_1 2^{\not{6}} \not{a}={}^7C_2 \not{2^5} a^{\not{2}}$
   $\mathbf{a=\cfrac{2}{3}}$
    
2. Assume $\theta=\tan^{-1}(3)$, Then $t=\tan{\theta}=3$
   $\sin^{-1}(x-1)=\tan^{-1}3$
   $(x-1)=\sin{\theta}$
   $(x-1)^2=\sin^2{\theta}$
   $=\cfrac{1-\cos 2\theta}{2}$
   $=\cfrac{1-\cfrac{1-t^2}{1+t^2}}{2}$
   $=\cfrac{1-\cfrac{1-9}{1+9}}{2}$
   $=\cfrac{18}{20}$
   $\therefore x-1=\pm \cfrac{3}{\sqrt{10}}$
   $\mathbf{x=1\pm\cfrac{3}{\sqrt{10}}}$
   *The answer can be easily taken if you are using table or calculators.
    
3. $\cfrac{13\sin^2\theta}{2+\cos\theta}+\cos\theta=2$
   $13\sin^2\theta=(2-\cos\theta)(2+\cos\theta)$
   $13\sin^2\theta=4-\cos^2\theta$
   $12\sin^2\theta+\sin^2\theta+\cos^2\theta=4$
   $12\sin^2\theta=4-1$
   $\sin\theta=\pm \cfrac{1}{2}$
   $\mathbf{\theta=30^0}$ or $\mathbf{\theta=150^0}$
    
4.   
1. As $4x+ky=20$ passes through $A(8,-4)$ and $B(b,2b)$,
   $4\times 8+k\times(-4)=20$
   $\mathbf{k=3}$
   $4\times b+3\times (2b)=20$
   $\mathbf{b=2}$
    
2. If the mid point is $M$,
   $M=\left(\cfrac{8+2}{2},\cfrac{-4+4}{2}\right)$
   $\mathbf{M=(5,0)}$
5. If the line $y_1=2x-k$ meets the curve $y_2=x^2+kx-2$ at two distinct points, then the quadratic equation $y_1=y_2$ must have two distinct roots. For that, the discriminant $\Delta _x$ must be positive.
   $2x-k=x^2+kx-2$
   $x^2+(k-2)x+k-2=0$
   $\therefore \Delta _x=(k-2)^2-4\times1\times(k-2)$
   But $\Delta _x> 0$
   $\therefore (k-2)^2-4(k-2)> 0$
   $(k-2)(k-2-4)>0$
   $(k-2)(k-6)>0$
   $\therefore$ either $k>2$ and $k>6$ or $k<2$ and $k<6$
   So the set of values of $k$ is $\mathbf{k>6}$ or $\mathbf{k<2}$
   
   
    
6.  
1. Here I show two methods
   Method 01
   $\overrightarrow{OA}=3i+2j-k$ and $\overrightarrow{OB}=7i-3j+k$
   $\therefore \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}$
   $=-3i-2j+k+7i-3j+k=4i-5j+2k$
   If $OAB\measuredangle=90^0$ then the scalar product between $\overrightarrow{OA}$ and $\overrightarrow{AB}$ must be zero.
   $\overrightarrow{OA}\cdot\overrightarrow{AB}=(3i+2j-k)\cdot(4i-5j+2k)$
   $=12-10-2=\mathbf{0}$
   
   Method 02
   If $\hat{OAB}$ is a right angle, then $OAB$ triangle must be a right angle triangle. So we must be use Pythagorean theorem for the triangle.
   $|\overrightarrow{OA}|^2=3^2+2^2+(-1)^2=14$
   $|\overrightarrow{OB}|^2=7^2+(-3)^2+1^2=59$
   $|\overrightarrow{AB}|^2=4^2+(-5)^2+2^2=45$
   $\therefore$ We can see that,
   $|\overrightarrow{OB}|^2=|\overrightarrow{OA}|^2+|\overrightarrow{AB}|^2$
   Hence, according to the converse of the Pythagorean theorem we can say $OAB$ is a right angle triangle.
    
2. Area $=\cfrac{1}{2}\times|OA|\times|AB|=\cfrac{1}{2}\times\sqrt{14}\times\sqrt{45}$
   $\therefore$ Area $=\cfrac{3}{2}\sqrt{70}$
    
7.  
1. According to the given data,
   $S=\cfrac{a\left(1-r^{\infty}\right)}{1-r}$  --------- (1)
   $3S=\cfrac{a\left(1-(2r)^{\infty}\right)}{1-2r}$ ------- (2)
   By dividing (2) by (1),
   $3=\cfrac{1-r}{1-2r}$    $(\because r^{\infty},$ $(2r)^{\infty} \to 0)$
   $3-6r=1-r$
   $\therefore r=\cfrac{2}{5}$
    
2. According to the given data,
   $T_1=7$, $T_n=84$ and $T_{3n}=245$
   If the common difference is $d$, then the general term is,
   $T_n=a+(n-1)d$  where $a$ is the first term.
   $84=7+(n-1)d$
   $(n-1)d=77$ -------- (A)
   $T_{3n}=a+(3n-1)d$
   $245=7+(3n-1)d$
   $(3n-1)d=238$ -------- (B)
   Dividing (B) by (A),
   $\cfrac{3n-1}{n-1}=\cfrac{238}{77}$
   $11(3n-1)=34(n-1)$
   $\therefore \mathbf{n=23}$
    
8.  
1. When we consider the given figure $OA=4 cm$
   By resolving the side,
   $OD=4\cos\alpha$ $cm$
   $\therefore$ The arc length $CD=4\alpha\cos\alpha$   $(\because A=r\theta)$
   $\therefore$ Total perimeter of $ABCD=4\alpha\cos\alpha+4\alpha+2(4-4\cos\alpha)$
   $=4\alpha(1+\cos\alpha)+8(1-\cos\alpha)$ $cm$
    
2. Area of a sector $=\cfrac{1}{2}r^2\theta$
   $\therefore$ The area of the shaded region $=\cfrac{1}{2}4^2\alpha-\cfrac{1}{2}(4\cos\alpha)^2\alpha$
   $=8\cdot\cfrac{\pi}{6}\left(1-\left(\cfrac{\sqrt{3}}{2}\right)^2\right)$
   $=\cfrac{1}{3}\pi$ $cm^2$
    
9.  
1. As $\mathrm{f}'(x)=2x-\cfrac{2}{x^2}$ we can get the gradient of the tangent ($m_T$) at $P(2,6)$,
   $m_T=2(2)-\cfrac{2}{2^2}=\cfrac{7}{2}$
   $\therefore$ the gradient of the normal $m_N$ satisfies,
   $m_T m_N=-1$  ($\because$ they are perpendicular to each other)
   $\therefore m_N=-\cfrac{2}{7}$
   As the normal passes through point $P$,
   $\cfrac{y-6}{x-2}=\cfrac{-2}{7}$
   $\therefore$ the equation of the normal is $\mathbf{7y+2x-46=0}$
    
2. By integrating w.r.t $x$, $\int \mathrm{f}'(x) \: \mathrm{d}x=\mathrm{f}(x)$
   
   $\mathrm{f}(x)=2\cdot\cfrac{x^2}{2}-2\cdot\cfrac{x^{-2+1}}{-2+1}+c$  Where $c$ is a constant
   $\mathrm{f}(x)=x^2+\cfrac{2}{x}+c$
   As $y=\mathrm{f}(x)$ passes through $P$,
   $6=2^2+\cfrac{2}{2}+c$
   $\therefore c=1$
   Then $\mathrm{f}(x)=x^2+\cfrac{2}{x}+1$
    
3. At stationary points $\mathrm{f}'(x)=0$
   $\therefore 2x-\cfrac{2}{x^2}=0$
   $x^3-1=0$
   $(x-1)(x^2+x+1)=0$
   As $x^2+x+1\ne 0$, $x-1=0$
   Therefore there is a stationary point at $\mathbf{x=1}$
   If you substitute two values in positive and negative sides of $x=1$, to the differential $\mathrm{f}'(x)$, you will find that this is a minimum. or
   differentiate $\mathrm{f}'(x)$ w.r.t $x$ again and substitute $x=1$, you'll get a positive value.
10.  
1. $x^2-2x-15=x^2-2x+\left(\cfrac{-2}{2}\right)^2-\left(\cfrac{-2}{2}\right)^2-15$
   $=(x^2-2x+1)-16=(x-1)^2-16$
    
2. $\mathrm{f}(x)=(x-1)^2-16$
   $c$ is the minimum value that can be taken for $\mathrm{f}(x)$
   $\therefore \mathrm{f}_{min}(x)=0-16=\mathbf{-16}\quad(\because 0$ is the minimum value for a quadratic$)$
   $c_{min}=-16$
    
3. If $\mathrm{f}(x)\to c$ then $x\to p$
   $9=(p-1)^2-16$
   $(p-1)^2=25$
   $p-1=\pm 5$
   $p=\mathbf{6}$  as $p$ is a positive constant
   $65=(q-1)^2-16$
   $(q-1)^2=81$
   $q-1=\pm 9$
   $q=\mathbf{10}$ as $q$ is positive.
    
4. $y=\mathrm{f}(x)=(x-1)^2-16$
   $(x-1)^2=y+16$
   $x-1=\pm \sqrt{y+16}$
   $x=1\pm \sqrt{y+16}$
   $0<x$ and $9<y$
   $\therefore \mathrm{f}^{-1}(x)=1+\sqrt{x+16}$
    
11.  


1. By differentiating both curves w.r.t $x$, at the point $Q(2,3)$, we can get the gradient of to tangents.
   $m_1=\left.\cfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{4x+1}\right)\right |_{x=2}=\left.\cfrac{1}{2}(4x+1)^{\frac{1}{2}-1}\cdot 4\right |_{x=2}$
   $m_1=\cfrac{2}{\sqrt{4\times 2+1}}=\cfrac{2}{3}$
   $m_2=\left.\cfrac{\mathrm d}{\mathrm d x}\left(\cfrac{1}{2}x^2+1\right)\right |_{x=2}=\left. \cfrac{1}{2}\cdot 2x\right |_{x=2}$
   $m_2=2$
   
   $\theta=\theta _2-\theta _1$
   $\tan\theta=\tan(\theta_2-\theta_1)$
   $\tan\theta=\cfrac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1}$
   $\tan\theta=\left|\cfrac{m_2-m_1}{1+m_2 m_1}\right|$
   $\theta=\tan^{-1}\left|\cfrac{2-\cfrac{2}{3}}{1+2\cdot\cfrac{2}{3}}\right|$
   $\theta=\tan^{-1}\left(\cfrac{4}{7}\right)=29.745^0$
    
2. Area $=\int\limits_0^2 y_1-y_2 \: \mathrm{d}x$
   $=\int\limits_0^2 \sqrt{4x+1}-\left(\cfrac{1}{2}x^2+1\right)\:\mathrm{d}x$
   $=\left[\cfrac{(4x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\cdot\cfrac{1}{4}-\cfrac{1}{2}\cfrac{x^3}{3}-x\right]_0^2$
   $=\left[\cfrac{(4x+1)^{\frac{3}{2}}}{6}-\cfrac{x^3}{6}-x\right]_0^2$
   $=\left\{\cfrac{(4\times2+1)^\frac{3}{2}}{6}-\cfrac{2^3}{6}-2\right\}-\left\{\cfrac{(0+1)^\frac{3}{2}}{6}-0-0\right\}$
   $=\cfrac{9^{^3/_2}}{6}-\cfrac{8}{6}-2-\cfrac{1}{6}$
   $=1$ unit
   

Stay connected for the next paper....

Centers of a Triangle

Triangle is an important polygon when we consider the geometry. It has several properties which are very specific to the triangle. In this article I have mentioned about the centers of a triangle.

1. Centroid
   
   
   
   Centroid is the point that the three medians of triangle meet together (concurrent). Following figure shows the geometry of the centroid. (median is the line, which joins the vertex and opposite mid point)
   
   Centroid is the center of gravity of a triangle and we can discuss it in the lesson "Center of Gravity".
   Here $AG:GE=BG:GF=CG:GD=2:1$
   $\therefore GE=\cfrac{1}{3}AE$
2. Circumcenter 
   
   
   Cercumcenter is the center of the circumscribed circle of a triangle. Concurrent perpendicular bisectors of each side of a triangle gives the circumcenter. 
   
   
   Here $AM=BM=CM=$ Radius of circumcircle. The angle subtend the center is equal to the double of the angle which subtend the circumference. So $AMB\measuredangle=2ACB\measuredangle=2C$.
   
   
3. Incenter
   
   
   Incenter is the center of the inscribed circle of a triangle. This is the concurrent point of three angle bisectors. The circle drawn with incenter as the center and the perpendicular length from incenter to the sides(which is equal) as the radius, is called the incircle.
   
   $ID=IE=IF=$ Radius of incircle.
4. Orthocenter
   
   
   Orthocenter of a triangle is the concurrent point of three altitudes. (altitude is the line drawn from a vertex to the opposite side, perpendicularly). Refer the following figure.
   
5. Nine point circle
   Nine point circle is the circle which goes through three mid points of sides, three feet of altitudes and three mid points of line segments from orthocenter to the vertices.
   
   Radius of circumcircle is two times the radius of nine point circle.

Note: There are other centers and lines (such as Euler line) specific to the triangle. Here I have described most important properties.

Trigonometry 02 (Measurements of Angles & Trigonometric Ratios)

• Definition of Angle

Let's assume that the line $OP$ is rotating in the same plane, around the point $O$ as shown in the figure. Let its origin is $OA$ and it is getting on to the positions $OB, OC, OD...$ respectively. Then we measure the angle between $OA$ and any position of $OP$ such as $OB$, using the amount of rotation that it made during its relocation from $OA$ to $OB$. According to this, there can be any number of complete rotations through $OA$, when $OP$ comes to $OB$. 

So not as in Geometry, in Trigonometry there can be any magnitude for the angles, without limitations. According to the above figure $O$ is the Origin, $OA$ is the Initial Line and $OP$ is the Rotating Line (Generating Line/ Radius Vector).

• Measurement of angles

Measuring the angles has happened in ancient history as well. For the measurement of angles we use the unit of complete revolution; that is the angle of a complete cycle which subtend the center of a circle. This complete angle has divided in to 360 equal parts and has called them degrees, by the ancient astronomers.

• Sexagesimal measurement 

In this measuring system, angle in a complete cycle has divided into four equal parts and has named them as right angles. Each right angle has divided into 90 equal parts and has introduced each part as a degree, and again each degree has divided into 60 equal parts and called each part as a minute. Finally each minute has divided again into 60 equal parts and named them as seconds.
The angle which describe as 54 degrees, 45 minutes and 3.45 seconds, can be denoted by $54^0$ $45'$ $3.45''$.

• Radian measurements

Practically we use the above mentioned sexagesimal system for measuring angles. But in theoretical applications radians are used widely for measuring angles. A radian has defined as the angle which subtend the center by an arc, having the length of the radius of the particular circle.

$AOB\measuredangle=1rad$

• The ratio between the circumference and the diameter, is a constant in any circle. 

To derive this, we have to consider the circle as a polygon with $n$ number of sides. Consider two circles with radii $r_1$ and $r_2$ and centers $O_1$ and $O_2$.

If we consider these two are regular polygons and assume the length of a side of the first polygon is $A_1 B_1$ and the length of a side of the second polygon is $A_2 B_2$.
So $A_1 O_1 B_1\measuredangle=A_2 O_2 B_2\measuredangle=\cfrac{360^0}{n}$.
Hence both triangles $A_1 O_1 B_1$ and $A_2 O_2 B_2$ become equi-angular.

$\therefore \cfrac{A_1B_1}{O_1A_1}=\cfrac{A_2B_2}{O_2A_2}$

Let $A_1B_1=l_1$ and $A_2B_2=l_2$, then,

$\cfrac{l_1}{r_1}=\cfrac{l_2}{r_2}$

$\therefore \cfrac{nl_1}{r_1}=\cfrac{nl_2}{r_2}$

If the number of sides $n$ is going to be increased infinitely, then $nl_1$ tends to be the circumference of the circle with radius $r_1$. Same thing happens to the other circle as well. If the circumferences of circles are $C_1$ and $C_2$ respectively then,

$\cfrac{C_1}{r_1}=\cfrac{C_2}{r_2}$

This shows us that the ratio between the circumference and the radius of any  circle is a constant. If so, the ratio between the circumference and the diameter of any circle is probably a constant.
$\therefore \cfrac{circumference}{diameter}$ is a constant.
This constant is represented by the Greek letter $\pi$. The value of this constant is approximately equal to the $\cfrac{22}{7}$. However exact constant is an Irrational number.

Circumference $C=2\pi r$ 

• Radians and Degrees

As we know the angle that  subtend the center of a circle by a complete cycle is $360^0$. According to the definitions of a radian,
Subtend angle by a $r$ arc length $=1$ $rad$
$\therefore$ The angle of a complete cycle $=\cfrac{1}{r}2\pi r$     $(\because C=2\pi r)$
$\therefore 360^0=2\pi$ $rad$
$1$ $rad=\cfrac{360^0}{2\pi}\approx 57^0 18'$

• Length of an arc that subtends the center by a $\theta$ angle.

To be continued...

Trigonometry 01 (Useful Formulas in Trigonometry)

1. $S=r\theta$  [Arc = Radius $\times$ Angle in radians]
2. Area of a sector $A=\frac{1}{2}r^2\theta$
3. Fundamental identities of trigonometry
• $\cos^2\theta+\sin^2\theta=1$
• $1+\tan^2\theta=\sec^2\theta$
• $\cot^2\theta+1=cosec^2\theta$
4. 
   

   $\theta$	$0^0$	$30^0$	$45^0$	$60^0$	$90^0$	$180^0$
   $\sin \theta$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$	$0$
   $\cos \theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$	$-1$
   $\tan \theta$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	$\infty$	$0$

    
5. $\sin(90^0-\theta)=\cos \theta$
   $\cos(90^0-\theta)=\sin \theta$
    
6. $\sin(-\theta)=-\sin \theta$
   $\cos(-\theta)=\cos \theta$
    
7. $\sin(180^0-\theta)=\sin \theta$
   $\cos(180^0-\theta)=-\cos\theta$
    
8. $\sin(90^0+\theta)=\cos \theta$
   $\cos(90^0+\theta)=-\sin \theta$
    
9. $\sin(180^0+\theta)=-\sin \theta$
   $\cos(180^0+\theta)=-\cos\theta$
    
10. $\sin(A\pm B)=\sin(A)\cos(B)\pm \cos(A)\sin(B)$
    $\cos(A\pm B)=\cos(A)\cos(B)\mp \sin(A)\sin(B)$
     
11. $\color{green}{\sin(2A)=2\sin(A)\cos(A)}$
     
12. $\color{green}{\cos(2A)=\cos^2(A)-\sin^2(A)=2\cos^2(A)-1=1-2\sin^2(A)}$
     
13. $\color{green}{\tan(2A)=\cfrac{2\tan(A)}{1-\tan^2(A)}}$
     
14. $\color{blue}{\sin(3A)=3\sin(A)-4\sin^3(A)}$
     
15. $\color{blue}{\cos(3A)=4\cos^3(A)-3\cos(A)}$
     
16. $\color{blue}{\tan(3A)=\cfrac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}}$
     
17. $\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2} \right)$
     
18. $\sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$
     
19. $\cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$
     
20. $\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)$
     
21. $2\sin{ }A \cos{ }B=\sin(A+B)+\sin(A-B)$
22. $2\cos A\sin B=\sin(A+B)-\sin(A-B)$
23. $2\cos A\cos B=\cos(A+B)+\cos(A-B)$
24. $2\sin A\sin B=\cos(A-B)-\cos(A+B)$
     
25. If $\sin\theta=\sin\alpha$, then $\theta=n\pi+(-1)^n\alpha$
    If $\cos\theta=\cos\alpha$, then $\theta=2n\pi\pm\alpha$
    If $\tan\theta=\tan\alpha$, then $\theta=n\pi+\alpha$
    Here $n\in\mathbb{Z}$
     
26. For an $ABC$ triangle, in the standard notation,
1. $\cfrac{\sin A}{a}=\cfrac{\sin B}{b}=\cfrac{\sin C}{c}$
2. $\cos A=\cfrac{b^2+c^2-a^2}{2bc}$
3. $a=b$ $\cos C+c$ $\cos B$
4. $\tan\left(\cfrac{B-C}{2}\right)=\left(\cfrac{b-c}{b+c}\right)\cot\cfrac{A}{2}$
5. $\cos\cfrac{A}{2}=\sqrt{\cfrac{s(s-a)}{bc}}$
   Here $s$ is the double of perimeter, of the $ABC$ triangle $(2s=a+b+c)$