Forward force is 50,000 and the resistive force is also 50,000 N which makes the total resultant force equal to zero.
At the time when acceleration begins the speed is 10 ms^-1 .
How can the acceleration be 0.20ms^-2 when the speed is still 10ms-1?
I just gave my phy paper2 in oct/nov session . It had a non rectilinear motion question in which we had to calculate max vertical displacement. I made a blunder and took the value of accelaration positive (9.81). It was of three marks i guess. the next question asked KE/GPE. so it subsequently used the height calculated in previous question . I took the other things completely fine.this was of 4 marks. Can anyone guess how much marks will I be losing?
The philosophy of reaching a maximum, any maximum, means that philosophically one is climbing and climbing higher and higher to reach the maximum peak, as long as there is a higher peak one keeps on climbing the slope. Once one reaches the top, the top must be a horizontal plane, even if it is a sharp peak, it must be a plane otherwise one has not reached the maximum.
Then one goes down the peak.
So, it is the same with velocity, if the velocity reached is a maximum, one can only go down the slopes around it, after retaining that maximum velocity for some time.
When acceleration is equal to zero than, F =ma a= 0 than resultant force is equal to zero.
As I have shown in the graph above although the acceleration begins to be 0.20 but the initial will remain 10 ms^-1 . the point where it begins it will change after wards.
has anyone got model diagrams for q1 of p5?
You will require those 3 subjects. Check the IBCC website though since they're the ones that create the Equivalency Form(you may get some leeway if you are a dual national).
units are same thus, both are equal.