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M1 i and j vectors
Hey, any chance someone could help with this questions, i'm finding i and j vectors really hard so any help would be great thanx.
Q
A particle P of mass 3kg is moving under the action of a constant force F newtons. At t=0, P has velocity (3i - 5j)ms-1 . At t=4s the velocity of P is (-5i +11j).
Find the magnitude of F?
Q
A particle P of mass 3kg is moving under the action of a constant force F newtons. At t=0, P has velocity (3i - 5j)ms-1 . At t=4s the velocity of P is (-5i +11j).
Find the magnitude of F?
Lettuce
Hey, any chance someone could help with this questions, i'm finding i and j vectors really hard so any help would be great thanx.
Q
A particle P of mass 3kg is moving under the action of a constant force F newtons. At t=0, P has velocity (3i - 5j)ms-1 . At t=4s the velocity of P is (-5i +11j).
Find the magnitude of F?
Q
A particle P of mass 3kg is moving under the action of a constant force F newtons. At t=0, P has velocity (3i - 5j)ms-1 . At t=4s the velocity of P is (-5i +11j).
Find the magnitude of F?
v=u+at
a=(v-u)/t
a=(-5i+11j)-(3i-5j)/4
a=(-8i+16j)/4
a=(-2i+4j)
F=ma
F=3(-2i+4j)
F=-6i+12j
Maginitude of F= √((-6)²+12²)
.....................=√(36+144)
.....................=√180
.....................=13.4N
I think thats the way you go about it.
Well the change in velocity is (-8i + 16j) (you just work out the difference in velocities).
Then divide by 4s to find the acceleration:
= (-2i + 4j)
Use pythagoras to find the magnitude of the acceleration:
|a|² = -2² + 4²
|a|² = 20
|a| = 4.47ms^-2
Then use F = ma to find the force:
F = 3kg x 4.47ms^-2
F = 13.41N
[EDIT] Damn someone always gets there before me. Oh well at least we get the same answers.
Then divide by 4s to find the acceleration:
= (-2i + 4j)
Use pythagoras to find the magnitude of the acceleration:
|a|² = -2² + 4²
|a|² = 20
|a| = 4.47ms^-2
Then use F = ma to find the force:
F = 3kg x 4.47ms^-2
F = 13.41N
[EDIT] Damn someone always gets there before me. Oh well at least we get the same answers.
Thanx
is there any chance you could help with the second part of the question only i'm still finding it really hard.
At t=6s P is at the point A with the position vector (6i - 29j)m relative to a fixed origin O. At this instant the force F newtons is removed and P then moves with constant velocity. Three seconds after the force has been removed, P is at the point B.
Calculate the distance of B from O?
is there any chance you could help with the second part of the question only i'm still finding it really hard.
At t=6s P is at the point A with the position vector (6i - 29j)m relative to a fixed origin O. At this instant the force F newtons is removed and P then moves with constant velocity. Three seconds after the force has been removed, P is at the point B.
Calculate the distance of B from O?
From the first part you have that a = (-2i + 4j), and at t = 0, u = (3i - 5j).
Using v = u + at,
You find that at t = 6,
v = (3i - 5j) + 6(-2i + 4j)
v = -9i + 19j
We have just been told that when t = 6 it is at A, which is 6i - 29j. The velocity is constant from this point on. It keeps going for 3 seconds before it reaches B.
s = ut = (-9i + 19j) x 3 = -27i + 57j
Add this on to its position at A gives: (6i - 29j) + (-27i + 57j) = -21i + 28j
The distance OB is therefore: root(212 + 282) = 35 m
Using v = u + at,
You find that at t = 6,
v = (3i - 5j) + 6(-2i + 4j)
v = -9i + 19j
We have just been told that when t = 6 it is at A, which is 6i - 29j. The velocity is constant from this point on. It keeps going for 3 seconds before it reaches B.
s = ut = (-9i + 19j) x 3 = -27i + 57j
Add this on to its position at A gives: (6i - 29j) + (-27i + 57j) = -21i + 28j
The distance OB is therefore: root(212 + 282) = 35 m
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