Thread: [SOLVED] mechanics

Can anyone help please?

A particle P starts from rest from a point A and moves alonga straight line with a constant acceleration of 2ms
at the same time, a second particle Q is 5m behind A and is moving in the same direction as P with a speed 5ms
if Q has a constant acceleration of 3ms find how far from A it overtakes P

Originally Posted by divine_mail

Can anyone help please?

A particle P starts from rest from a point A and moves alonga straight line with a constant acceleration of 2ms
at the same time, a second particle Q is 5m behind A and is moving in the same direction as P with a speed 5ms
if Q has a constant acceleration of 3ms find how far from A it overtakes P

start by making a table
t p q
0 0 -5
1 2 2
2 6 12
3 12 25

this gives you an idea of whats happening. you can see that you can make an equation for both particles

we know that acceleration for p, or p'' is:
p''=2 so we can get
p' = 2x + c where c represents initial speed, in this case...
p' = 2x + 0 (it started at rest)
then we find its position
p = x^2 + c where c=0 (or at least 5 more than c in q(x))

then do the same for q
q''= 3
q' = 3x + c (initial speed was 5m/s)
q' = 3x + 5
q = (3/2)x^2 + 5x + c (initial positon was 5 behind p)
q = (3/2)x^2 + 5x - 5

now set p and q equal to eachother and that will tell you the moment they intersected.

Originally Posted by divine_mail

Can anyone help please?

A particle P starts from rest from a point A and moves alonga straight line with a constant acceleration of 2ms
at the same time, a second particle Q is 5m behind A and is moving in the same direction as P with a speed 5ms
if Q has a constant acceleration of 3ms find how far from A it overtakes P

And please note that the acceleration is given in units of m/s^2, not ms.

-Dan