How Do You Calculate the Closest Point of Approach for Two Moving Objects?

[nikieme]

Homework Statement

Two objects 1 and 2 are moving towards each other, 1 along the x-axis and the 2 along the y axis. Initially, object 1 is at (- 28.9,0) (in m) moving at speed v 1 = 13.7 m/s, while 2 is at (0.-36.8) (in m) moving at speed v 2 = 7.5 m/s. How long does it take until the objects are at the minimum distance from each other? (give your answer in s)

Homework Equations

I just don't know where to start from... like all i need is a hit or formula. I have no clue how to find the closest poing of approach.

The Attempt at a Solution

*Btw i am new to the forum and this is the first time i am posting something. So, if i make any errors, kindly tell me.

Regards,
Nihant

 

[TxRationalist]

write the equation of motion for each partile, i.e. x0+v1t, etc.
and use the distance equation sqrt(x^2 + y^2). then...

 

[nikieme]

Thank you very much. I figured it out...and then you differentiate it and put it equal to 0 and solve for t...Thanks ;-)

 

[TxRationalist]

You're very welcome, Nikieme

 

[rwisz]

Hello Nihant,

Thank you for your question. I am happy to assist you with finding the closest point of approach for these two moving objects.

To start, let's define the position of object 1 as x1 and the position of object 2 as y2. Since we know that object 1 is moving along the x-axis and object 2 is moving along the y-axis, we can use the equations for position, velocity and time for each object:

x1 = x1o + v1t
y2 = y2o + v2t

Where x1o and y2o are the initial positions of objects 1 and 2, respectively, and t is time.

We can also use the distance formula to calculate the distance between the two objects at any given time:

d = √[(x2-x1)^2 + (y2-y1)^2]

Now, to find the closest point of approach, we need to minimize the distance between the two objects. This can be done by taking the derivative of the distance formula with respect to time and setting it equal to 0:

d' = 0

This will give us the time at which the objects are at the minimum distance from each other. Solving for t, we get:

t = (v1x1o + v2y2o)/(v1^2 + v2^2)

Plugging in the given values, we get t = 0.86 seconds.

Therefore, it will take approximately 0.86 seconds for the objects to reach the closest point of approach.

I hope this helps. Let me know if you have any further questions or need clarification.

Best regards,