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…
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.
The equation of motion applicable when the origin of time is zero are
After 3 seconds , X will have reached the position and speed
Sx = 2.8*3+1/2*0.12*3^2 = 4.54
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
whose meaningful solution is t=27.2s , corresponding to a distance of 139.4m
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
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.
The sequence is
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%
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%