Discussion on Kinematics


 

Join Now for Exam PreparationSTART YOUR BITSAT PREPARATION
  •  More than 1 lakh Students
  •  Practice 1000s of questions
  •  Chapter-wise Study Material
  •  Guided Study Plan
WHAT ARE YOU WAITING FOR? JOIN NOW

Posted on: 03:22:19 PM on April 19, 2016
Posted By:
AJAY SHARMA
A pedestrian is running at his maximum speed of 6. m/s trying to catch a bus that is stopped at a traffic light. When he is 16 m. from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1. m/s/s.


REGISTER NOW & POST YOUR QUESTIONS
 

Reponses To This

Posted on: 03:23:42 PM on April 19, 2016
a) Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian's distance of closest approach?) b) How fast is the bus moving when the pedestrian catches it? ( or at the distance of closest approach) I think I know that for the pedestrian to catch the bus, the displacements have to be the same, along with time. The pedestrian also has to make up for the 16m. so his displacement would be d+16....but I'm stuck, I don't know how to answer the question. Please Help. Thanks



Posted on: 08:05:11 PM on April 25, 2016
Suppose pedestrian and bus are at same point after the bus has traveled 'X' distance. Time to cover distance 'X'--> X=Ut+1/2at^2. U=, a=1; t=sq. root of 2X Now, pedestrian is not accelerating, therefore, speed=X/t; 6=X+16/(2X)^1/2. Solving this quadratic equation gives X=-8,-32. This shows that man can not catch bus. Therefore, the closest distance of approach is 8 metres which means when the pedestrian travels 8m bus would be closest but after that distance would start to increase. I think your doubt has been finished till here and you are intelligent enough to find the velocity at this moment. Hope it Helps. PS: BEST OF LUCK MY FRIEND. HOPE WE MEET AT PILANI CAMPUS.





Post Your Response




 
SimplyLearnt
Sign Up to participate in this discussion