Again are you sure about this? Because the direction is changing, so therefore the graph will not be straight (ie acceleration).
However when resolved into x and y coordinates, then yes it would be straight.
Anything wrong with this thinking??
I remember that anything in orbit is in...
Hopefully I'm posting in the right section of the forum. I want to find out how the velocity-time graph (or any other graph for that matter) will look like for a projectile.
Because I want to understand the theory behind projectiles properly and I can't seem to find any answers online...
One problem though. In the the a) part of question 6 it asked:
"Question 6: The force X in Question 5 is replaced by a force Y at an angle of 45 degrees to the horizontal. Find Y when the body is on the point of slipping
a) up the plane."
And I basically used (-45 degrees). And I got...
Yeah I choose the one that is points downward to the right. By anticlockwise, I mean the angle the line makes with the horizontal. Going anti-clockwise until it's 45 degrees.
But I simply added a 45 degree angle to X force. My problem is understanding the question I guess, where am I going...
Okay here's the first question which I done:
"Question 5: A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling...
Okay going to be a bit difficult as you need a force diagram. But just imagine a box on an inclined plane 20 degrees to the horizontal, with mg acting down, Normal Reaction and Frictional Force.
Therefore Frictional Force=18.4x0.2=3.68N
I'm not getting some mechanics questions. I'm sure that I'm right... but the answer in the back of the book is different. I'll type it out for you guys
"A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a...
Yes the quadratic formula bit was wrong.
As for the 2nd method, what I am trying to say is... the velocity=distance divided by time right?
Okay... so the distance the Man has to travel is 20 metres in addition to the distance the bus traveled in that time i.e. 1/2at^2 and the usage of the...
so you're saying that there's no solution?! lol
wow... I tried a few ways... and since V(of person) is constant, I just used that as a constant.
Let me show you:
Distance of Man= 20(metres)+1/2at^2=velocity x time
Rearranged it into: 3t^2-2vt+40=0
When you are 20m away from your bus it begins accelerating at 3 m/s/s (from rest). With what constant velocity should you run at to catch the bus?
The Attempt at a Solution
I've done loads of...