two cars start from the same point at the same time.the first car travells at constant speed of 12km/hr while second car travels at intial speed of 6km/hr,which increases by 1km/hr every hour.what will be the distance travelled by second car by the time it meets the first car

• First car speed intervals per hour
  12,12,12,12,12,..........n
  So sum of the speed's can be written as 12*n...............EQUATION 1
  Second car Speed intervals per hour
  6,7,8,9,10,11,12...........n
  So sum of the speeds would be
  Sn=n/2[2a+{n-1}d]
  Therefore sn=n/2[2*6+{n-1}1]
  Simplifying sn=n/2[12+n-1]........................Equation 2
  Equate equation 1 and 2 then we know where these equations would meet
  12n=n/2[12+n-1]
  Here do not cancel 12n with n/2 as we have to know the Value of n we shouldn't cancel it
  so bringing 12n to the other side
  n/2[12+n-1]-12n=0
  sending the denominator 2 to the other side it becomes zero and 12n becomes 24n as it is multiplied by 2....
  so the equation is n[12+n-1]-24n=0
  12n+n^2-n-24n=0
  n^2-13n=0
  n^2=13n
  n=13....
  Here as we taken the sum of the speed's per hour and equated them to know when do they meet
  now N mean after N hours those 2 equations meet so after 13hrs those two cars would meet
  And now to know the distance travelled by 2nd car to meet 1st car after 13hrs
  have to find the sum
  6,7,8,.................18
  sn=[first term+last term]/2 *n
  6+18/2*13
  24/2*13 therefore 12*13=156 is the ANSWER
• 9 years agoHelpfull: Yes(5) No(0)
• the distance travelled by second car will be 156km
• 10 years agoHelpfull: Yes(4) No(1)
• 156...Other than equations of motion..............
  car a distance pattern---12,24,36,48........................seq..1
  car B distance pattern--6,13,21,30,40.................. sequence 2
  A/q car b meets A, so seq 2 will be at a term divisible by 12.......And following seq -2 the first number is 156..(Brute --force)
  
• 10 years agoHelpfull: Yes(3) No(1)
• how,pls explain it
• 10 years agoHelpfull: Yes(1) No(0)
• dist. travelled by 2nd car =144 km. ans
  12*t=6t +1/2*1*t^2
  (,12*t=ut+1/2(a*t)^2)
  t(t-12)=0
  t=12
  total distance=12*12=144
  
• 10 years agoHelpfull: Yes(1) No(3)