How to tell how far a particle has travelled?

[hopbin9]

Hi,

If I have a particle with a velocity of v(0,1,0) then how many frames or seconds will it take to travel 1 unit of distance?

[Erik_JE]

1 m/s for 1 s makes it travel 1 m. So if your scene scale is m that would make it 1 unit in 1 s

[hopbin9]

thank you.

[gaurav]

thank you.

Hey hop ,thats right if the particle is in space may be on International Space Station.

But things are little different when the very same particle is on earth.

Lets say you had the particle gun and shot one from 1 meter above the ground plane with initial velocity of 30 m/s @ an angle of 30 degrees.

So the question is will it hit 1.75 meters tall, Bad guy (read me), 80 meters away from it ? Given the only force is acting is gravity g = -9.8m/s^2. and we are at the sea level.

For simplicity, lets discard dimension Z and assume we both are standing lines , living in xy flat land + you did not cheat and copy a small rock on it the moment it was born.

Calculating Position as a function of time and using unit vector notation.

we can express them analytically in n dimensions a = 6i+3j, b = -2i+2j, a+b = 4i+5j

So here it goes :

P(t) = P(initial) + V(initial).t + a.t^2/2 ------(1)

P(initial) = 0i+1j = 1j

V(initial) = 30cos30i + 30sin30j

V(initial)=30*.866 i + 30*.5 j

V(initial)= 25.98i + 15j

g = -9.8

a = 0*i- 9.8j a = -9.8j

replacing values in the equation 1

p(t) = 1j+(25.98i+15j)*t-9.8j*t^2/2

p(t)= (25.98*t)i + (1+15t-4.9t^2)j

Wohaaa..!!

we now know the x and y position at any given time

Solving for time when particle is 80 meters away. 25.98*t = 80 t = 80/25.98 = 3.07 secs.

Solving for y position at time 3.07 secs.

1+15*3.07-4.9*3.07^2

1+46.05-46.15

1-0.1 = .9 meters.

.9M < 1.75M " I will get hit exactly after 3.07secs".

hth,

Edited February 6, 2012 by vectorblur