TrueTears question thread

[shinny]

is simply the derivative of , i.e. . Is that what you were trying to find?

[TrueTears]

soz typo, i meant how do u know to use the function here to work out the angle

[TrueTears]

and also:

The velocity vector of a particle at time t seconds is given by . Find the magnitude and direction of the acceleration after one second.

I've worked out everything except for the direction part, how do u work out the direction of the acceleration?
My working is as follows:









Now im just stuck on how to work out the direction ><

[shinny]

Ah sorry, the function is of course what we refer to as the particle's velocity. Velocity not only indicates the speed at which something is moving, but the DIRECTION it is moving. The angle we are trying to find is the angle between the ground and it's initial direction. Hence, just draw up a triangle and find cos theta using the resolutes of the direction.

[TrueTears]

ahhhh that explains it all, thanks shinny.

[shinny]

and also:

The velocity vector of a particle at time t seconds is given by . Find the magnitude and direction of the acceleration after one second.

I've worked out everything except for the direction part, how do u work out the direction of the acceleration?
My working is as follows:









Now im just stuck on how to work out the direction ><

The direction of the acceleration is simply   Could convert this to a unit vector, but I don't think its necessary?

[TrueTears]

yeah.. that's what i thought as well, but the answer in the book i guess converted to unit vector, guess ill try that as well lol

[TrueTears]

1. A body moves horizontally along a straight line in a direction of with a constant speed of 20m/s. If is a horizontal unit vector due east and is a horizontal unit vector due north and it find the velocity of the body at time .

[/0]

I'm not sure about this, but since and is in quadrant 2, if you draw a diagram of that triangle then you'll see that the velocity vector in that direction is for some constant k.







So the velocity at a time t is

[TrueTears]

thx /0, that is the correct answer,

Also just this question


question 9 part a) just a bit confused.

[humph]

9 a) is asking for the resultant force as a vector, 9 b) is asking for the magnitude of that vector and then the unit vector.


So you're supposed to weight each of the forces correctly and then add them together. They've given you the direction of each vector, but they're not normalised. So the first thing to do would be to consider the three vectors when normalised; they are . Then you multiply these normalised vectors by the magnitude of the force of each one; . Then you add the three together; . So that's the resultant force. Then the magnitude is 5 and the direction is .

[TrueTears]

i kinda get it, but what do u mean by "normalised"? As in the unit vector in each of those 3 forces directions?

[ell]

i kinda get it, but what do u mean by "normalised"?

To normalise a vector is to take the unit vector.

[TrueTears]

and also Q 16 part a), Q 18 and Q 19. Just slightly confused ><""

[/0]

I am a noob in dynamics but whatever...

16a)

Let OX lie on the x-axis. The resultant force vector can be found by vector addition. Let the vector of 2N lie so that's its tail coincides with the head of the 3N vector. The resultant vector is the third side of the triangle.
The angle between the 2N and 3N vectors is

So the resultant vector has magnitude by the cosine rule.

From the vector triangle you can then use the cosine rule again to find the other angles, and hence the angle of the resultant vector with respect to the OX vector.

17. Same principle; draw a vector triangle

18. Draw a diagram and split the vectors into horizontal and vertical components, with the 10N vector on the x-axis. Since the resultant vector acts only along the 10N vector, the vertical vectors must cancel out, i.e.



.

Since P does not have a horizontal component, it does not contribute to the force along the 10N vector direction, so the resultant force along the 10N vector direction is: