Question

Particle A:
x(displacement) = (-t^2 + 4t +2)m
Particle B:
x(displacement) = (t^2-2t+8)m
v=(2t-2)m/s

Use algebra and calculus to show that the min distance between the two particles is 1.5m.

Answer 1

umm... anyone can do it?

Answer 2

Here is a graph of these functions,

http://www.wolframalpha.com/input/?i=x%3D-t^2+%2B+4t+%2B2+and+x%3D+t^2-2t%2B8

The time when the two functions are closest will occur at the time when the slope of the two functions are equal.

So if we look at the first equation, x is a function of t. When we take the derivative of

\(x=-t^2 + 4t +2\)

We end up with

\(x_{1}'=-2t + 4\)

And for the second equation:

\(x_{2}'=2t -2\)

Set them equal to each other then solve for t.

Then plug that t into each displacement equation. The difference between the two solutions should be 1.5 m

Answer 3

ok thx i ma try it

Answer 4

The time when the two functions are closest will occur at the time when the slope of the two functions are equal.

Does that apply to all functions ?

Answer 5

I believe it is true for two parabolic functions that are opposite in concavity and never touch.

Answer 6

ohh genius it worked :D

Answer 7

If i didnt had the graph I wouldn't have expected the minimum distance would occur when 2 slopes of the function are equal

Answer 8

Me either

Answer 9

There is surely another way to do this...

Answer 10

Thanks for ur help :D mind if u can help me with 1 more question? :D

Answer 11

sure

Answer 12

http://openstudy.com/study#/updates/503653f5e4b09d6877142b6c
its on this