Question

2 boys a and b move with constant velocities along two mutually perpendicular straight lines towards the intersection time at point P at time t=0.The boys were located at distance l1 and l2 respectively from P.then the time at which they are most near or equal to =?

Answer 1

This is a minimization problem. in this case d is to me minimized with respect to time. Write the equation of the distance d between the two persons in terms of time . Then take the derivative of this expression with respect to time .
Thus dd/dt =velocity of approach before they are closest and is positive. When dd/dt is negative they are receding from one another. They are closest when dd/dt =0 Solve this equation for time. If you need additional assistance let me know.

Answer 2

Actually, dd/dt will be negative on the approach (the distance between them is decreasing over time), and dd/dt will be positive on the departure (distance is increasing). I love problems like this; simple and easy to state, complicated and challenging to solve. I had to write two parametric equations in time, relate them by Pythagoras' theorem, then solve the minimization problem. The units seem to work out. Tell me if/how I messed up.

Answer 3

I looks good. At the end what was your intention for taking the derivative of time?

Answer 4

Cool, thanks. It's just confusing notation; d((a+b)/2v) isn't a derivative, it's the distance function (d(t)) evaluated at such time when dd/dt=0.

Answer 5

well..im taught this way..

Answer 6

For x to be minimum:
\[\LARGE \frac{dx}{dt}=0\]
\[\LARGE \frac{1(2(l1-v_{1}t)(-v_{1})+2(l_{2}-v_{2}t)(-v_{2})}{2 \sqrt{(l1-v_{1}t^{2})+(l_{2}-v_{2}t)^{2}}} =0\]

Answer 7

\[\LARGE t=\frac{l_{1}v_{1}+l_{2}v_{2}}{v_{1}+v_{2}} \]

Answer 8

@gleem @typingman

Answer 9

OOPs DLS your answer is not dimensionally correct.