General

[quote=MrPickles]We know that the cart comes to a stop over some distance.
We also know that due to Newton's first law, the total force applied to the cart over the 12m distance must bring the carts velocity to 0 m/s.
If this condition is not met, the cart will either start moving backwards due to excessive force or continue moving east due to the lack of it.
Therefore, the cart's final position must be 12m

I have no idea how to prove this mathematically but this was the conclusion I came to after some thinking.
IF YOU FEEL LIKE MY LOGIC IS OFF, I IMPLORE YOU TO QUOTE ME SO I CAN FIX THIS AND LEARN FROM THIS. >.<[/quote]

Never mind, I can't read.

Yeah, you have a really bad question OP. You actually should complain to your professor.

We know that the cart comes to a stop over some distance.
We also know that due to Newton's first law, the total force applied to the cart over the 12m distance must bring the carts velocity to 0 m/s.
If this condition is not met, the cart will either start moving backwards due to excessive force or continue moving east due to the lack of it.
Therefore, the cart's final position must be 12m

I have no idea how to prove this mathematically but this was the conclusion I came to after some thinking.
IF YOU FEEL LIKE MY LOGIC IS OFF, I IMPLORE YOU TO QUOTE ME SO I CAN FIX THIS AND LEARN FROM THIS. >.<

Edit: The math suggests that the shopping cart does not have a final position and that it will start moving west after 12m.

Since work (W) is equal to change kinetic energy (KE_change), the work done to get the cart to stop will be equal to 1/2 mv^2 = 121.5 J
However, since it is assumed that the applied force deceases steadily, i.e. uniformly, it would make sense that the equation for the applied for is F_a = 30-2.5x where x is the distance between 0 and 12m. Using the definition of work as an integral, we integrate F_a from 0 to 12m to get a value of -180J. Since I-180I > 121.5, it is implied that the cart will stop at some point, and start moving west to a position an infinite distance away.

Edit 2: This is all assuming that the applied force decays to 0N at the 12m mark. If this assumption is invalid, I assert that this problem is unsolvable since we do now know the rate of decay of the applied force.

[quote=HolyDragon]The last question as in question 3?

30N over 12m
On a graph, start at 30 position 0 and draw a straight line to 0 at position 12.
The area of that graph is the total applied force (Fa).

Fa=Applied Force
Ff=Friction Force

Fa-Ff=m*a
Find the total acceleration to find the velocity at position 12

Afterwards
Ff=m*a
Use a
Velocity at position 12 is Vi here
Vf=0
vf^2=vi^2 +2ad

Please check the above formula, I may have recalled it wrong.[/quote]

There is no frictional force.

The problem is also worded badly. It states that the force decreases steadily until the cart hits 12m. The problem doesn't state that the force [b]decreases to zero[/b], meaning that the cart would have a non-zero net velocity unless the force reduced its velocity to zero.

The last question as in question 3?

Edit: The guy below has indicated that I misread. Fixing

30N over 12m
On a graph, start at 30 position 0 and draw a straight line to Y at position 12.
The area of that graph is the total applied force (Fa).

Fa=Applied Force

-Fa=m*a

It has to have enough force that it stops at position 12
Rectangle + Triangle=ma

(Y*12)+(30-Y)6=ma

Might've messed up some signs somewhere.

its probably on google... just google it.

How's after school, it's easy but cbf writing it on basil in class

[quote=DeprivedChild]@iHartz: when do you need the help[/quote]

like right now

@iHartz: when do you need the help

[quote=DeprivedChild]First up what cs thing can you pay[/quote]

p250 cartel ft

First up what cs thing can you pay