# ERJ Brainteaser: March

20 Mar 2023

Each month, ERJ sets a weekly brainteaser, with questions of varying degrees of difficulty. Readers supplying the most accurate (and stylish) answers are then considered for the prestigious **Brainiac of the Month** title.

Question 4:** Keep it country II**

Fill in the gap in this sequence:

… Switzerland, Austria, __, Denmark, Norway, Poland…

Clues: 00

Email your answer: correct replies on Friday*.

**With apologies, for technical reasons the results of this week's brainteaser will now appear on Monday.*

Question 3: **Particle science**

Particle X passes through point P with velocity 2.8 m s−1 and constant acceleration 0.12 m s−2. Three seconds later particle Y passes through P with velocity 2.4 m s−1 and constant acceleration 0.2 m s−2. What is the distance from P when the two particles meet?

*Clues: Clues: This equation of motion might help: s = ut + 0.5 (a x t squared)(Where s = distance, u + initial velocity, v= velocity at time t, a = acceleration and t= time.)*

**Answer**: Our questions usually garner a handful of replies within a couple of hours. Not so with this one, which held out for almost a day before the first correct answer arrived in. Extra well done so, in order of reply, to: **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Trevor Elison**, manufacturing engineer, Mountville Rubber Co., LaGrange, Georgia, USA; **Andrew Knox**, Rubbond International, The Netherlands.

Solutions

*Michele Girardi*

*The equation of motion applicable when the origin of time is zero are *

*S=S0+V0*t+1/2*a*t^2*

*V=V0+a*t*

*After 3 seconds , X will have reached the position and speed *

*Sx = 2.8*3+1/2*0.12*3^2 = 4.54*

*Vx= 2.8+0.12*3=3.16*

*Let's take t=0 when Y is in P , the equations of motion are *

*X : S = 4.54+3.16*t+1/2*0.12*t^2*

*Y: S = 2.4*t+1/2*0.2*t^2*

*the time at which the particles will have the same position is given by *

*4.54+3.6*t+1/2*0.12*t^2=0+2,4*t+1/2*0.2^t^2*

*whose meaningful solution is t=27.2s , corresponding to a distance of 139.4m*

*Trevor Elison *

*Particles meet 140m from point P*

*Assume both particles are on the same path.*

*Assume both accelerations are measured with the same accuracy. (.2= .20 m/s^2)*

*Solve for s_x at 3.0 seconds = 8.94m**Solve for v_x at 3.0 seconds = 3.16m/s (v_x,3 = u_x + a_x*t)**Set s_x,f = s_y,f at time final (t_f)**Rearrange to format ax^2 + bx + c = 0*

*a = (a_y-a_x)/2, b = (u_y – v_x,3), c = -s_x,3 (.04, -.76, -8.94)*

*Solve for t_f using the quadratic formula = 27.22s (from the time Y passes point P)**Substitute t_f into either distance equation (at the time Y passes point P) to obtain distance where the two particles meet**139.3659m. Round for significant figures = 140m*

*Andrew Knox*

*Distance from point P when the two particles meet (assuming they are travelling in exactly the same direction!) is approx. 139.27 m, time elapsed for particle Y is approx 27.2 s after passing point P.*

*Calculation: Distance travelled by particle X in its 3 second "head start" is v.t + (1/2) a.t.t, so distance = 2.8.t + 0.06.t.t or 8.94 m.*

*Its velocity after 3 seconds is 2.8 + 0.12x3 = 3.16 m/s.*

*So the distance travelled from point P by X, from the moment Y passes point P, is Dx = 8.94 + 3.16T + 0.06 t.t*

*This needs to equate to the distance travelled from point P by particle Y which is Dy = 2.4.t + 0.1.t.t.*

*The two particles meet when Dx = Dy, so solve for t = ca. 27.2 s, and Dx & Dy are then both equal ca. 139.27 m.*

New, slightly easier, teaser on Monday.

Question 2: **What comes next?**

08 Friday, 12 Wednesday, 16 Monday, 20 Saturday, ?...

**Answer**: 24 Thursday ie the next ‘leap day’ on the calendar, 29th February 2024 – though some of our Brainiacs avoided all that and went straight down the maths route. Very well done, in order of reply, to: **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna; **David Mann**, key account manager, SPC Rubber Compounding, UK; **France Veillette**, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; **Suman Dhar**, deputy general manager- international business at zinc chemicals producer Rubamin Private Ltd, Vadodara, Gujarat, India; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **John Coleman**, membership manager, Circol ELT, Dublin, Republic of Ireland. And, an extra well done to those – you know who you are! – who took the leap-year route.

New teaser on Monday.

Question 1: **Percentage play**

Complete this sequence:

O = 10%, ONE H = 40%, ONE HUND = 70%, ___.

**Answer**: You either got this one quite quickly or, perhaps, went on a ‘wild goose chase’. As explained nicely by David Mann and Michele Girardi (see Solutions below) below, the answer is ONE HUNDRED = 100%. Very well done, in order of reply, to: **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **David Mann**, key account manager, SPC Rubber Compounding, UK; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, The Netherlands; **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Suman Dhar**, deputy general manager- international business at zinc chemicals producer Rubamin Private Ltd, Vadodara, Gujarat, India; **John Coleman**, membership manager, Circol ELT, Dublin, Republic of Ireland; **Serafina Vulcano**, Martinello Articoli Tecnici SpA, Italy; and everyone else who had a go.

Solutions

*David Mann*

* The sequence is*

*O*

*ONE*

*ONE HUND*

*ONE HUNDRED*

*and the percentages are the number of letters divided by the number of letters in ONE HUNDRED.*

*So the next term is ONE HUNDRED = 100%*

*Michele Girardi*

*Hello everybody, the answer is: ONE HUNDRED=100%, as can be seen from the following sequence: *

*O 1 letter 10%*

*ONE H 4 letters 40%*

*ONE HUND 7 letters 70%*

*ONE HUNDRED 10 letters 100%*