# Need help with simple position and velocity problems

One problem reads: A particle of mass 1 slug is moving in a constant force given as F= 3i + 10j - 5k lb
THe particle starts from rest at position (3,5,-4). What is the poisiton and velocity of the particle at time t= 8 sec.? WHat is the poistion wehn the particle is moving at a speed of 20 ft/sec?

I guess i am confused on how to find the position vector of teh particle at t= 8 sec. I know once i find that i need to subtract it from the orginal position. I then assuem i am going to do an integral? to get the position when the particle is moving at a speed of 20 ft/sec?

HEre is the second problem. I have no idea what to do. I am sure i need to take the derivative of the equation but i dont' know how to apply it.
http://www.mustangmods.com/data/16002/review1.jpg

Related Introductory Physics Homework Help News on Phys.org
for the second one (taking up and to the right to be positive)
Y Direction
v1 = 20 sin 60
v2 = ?
a = -9.8 m/s^2(gravity points downward
y = ?
X Direction
v1 = 20 cos 60
v2 = 20 cos 60 (No forces in the X direction)
a=0
x=?

and time = ?
Write down an equation for x in terms of what you are given for x. Do the same for y. Now how are x and y related, according to the diagram? Thus find the time, and you can find x, thus y and the point on the parabola.

I'll be more than happy to help you, but first I need to know what "moving in a constant force" means? I'm almost 100% sure that the problem has the force vector as the net force, meaning that the particle is accelearting. If this is the case, let me know and Ill help you get started. (BTW, yeah, there is a small deal of calc, but not to worry)

DaMastaofFisix said:
I'll be more than happy to help you, but first I need to know what "moving in a constant force" means? I'm almost 100% sure that the problem has the force vector as the net force, meaning that the particle is accelearting. If this is the case, let me know and Ill help you get started. (BTW, yeah, there is a small deal of calc, but not to worry)
yes, the particle is accelerating.