Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

suppose while sitting in a parked car you notice a jogger approaching towards you in a side view mirror of R=2m. if the jogger is running at a speed of 5m/s . how fast the iimage of the jogger appear to moove when the jogger is (a)39m (B)29m (C)19m (D)9m away. 

Asked by Kb Aulakh 6th March 2015, 7:51 PM
Answered by Expert
Answer:
begin mathsize 12px style Given colon Speed space of space jogger space from space the space mirror space equals space 5 space straight m divided by straight s Radius space of space curvature comma space straight C space equals space 2 space straight m semicolon space straight f space equals space straight c divided by 2 space equals space 1 space straight m Mirror space formula comma space 1 over straight f space equals space 1 over straight v space plus space 1 over straight u straight v space equals space fraction numerator fu over denominator straight u space minus space straight f end fraction for space straight u space equals space minus 39 space straight m straight v space equals space fraction numerator negative 39 space cross times space 1 over denominator negative 39 space minus 1 end fraction space equals space 39 over 40 space straight m since space the space jogger space moves space with space constant space speed space 5 space straight m divided by straight s. Hence comma space after space 1 space sec space the space jogger space would space move space 5 space straight m straight u space equals space 39 space minus space 5 space equals space 34 space straight m straight f space equals space 1 space straight m 1 over straight v space equals space 1 over straight f space minus space 1 over straight u space equals space 1 over 1 space minus space fraction numerator 1 over denominator negative 34 end fraction space straight v space equals space 34 over 35 space straight m Speed space of space the space image space of space jogger space equals space distance over time Therefore space the space shift space in space the space position space in space 1 space sec space is comma space left parenthesis change space is space velocity right parenthesis space cross times 1 space equals space shift space in space position open parentheses 39 over 40 space minus space 34 over 35 close parentheses space cross times 1 space equals space 5 over 1400 space equals space 1 over 280 space straight m Hence comma space average space speed space of space jogger space between space 39 space straight m space and space 34 space straight m space is comma straight v space equals space fraction numerator 1 over 280 over denominator 1 end fraction space equals space 1 over 280 space straight m divided by straight s Similarly space we space can space calculate space ofr space minus 29 space straight m comma space minus 19 straight m comma space and minus 9 straight m. space and space the space speed space og space joggers space would space be space 1 over 150 space straight m divided by straight s space comma space 1 over 60 space straight m divided by straight s space and space 1 over 10 space straight m divided by straight s space respectively. end style
Answered by Expert 7th March 2015, 9:38 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!