Recent content by doublev231

Looks like that's it then! Thank you very much once again for going through all of this with me! You've done me a huge favor. Thank you a ton!

Can I also write it like this: ... = |-7 m/s| = 7 m/s ? Also, if after 5 seconds it completely stops, then how can it have a velocity anymore? Shouldn't it be 0?

Oh, okay, I think this is the formula, right? v = u + at Therefore, v = 0 m/s + 2.8 m/s2 * 2.5s = 7 m/s Is that correct? The cart returned back to it's original speed? But does that make sense, because after 5 seconds, it's supposed to stop, so how can it have a velocity anymore?

s1 equals = 8.75 meters, s2 equals = 8.75 meters. In total I got 17.5 meters! Yay! Thank you so much! So, another task was final speed. Does this mean that the final speed will be equal to 0 meters per second?

Okay, this is what I tried to get the distance traveled in 2.5 seconds: t(to stop) = 7 / 2.8 = 2.5s s = 7m/s * 2.5s + 0.5 * (-2.8m/s2) * 2.5s2 = 17.5 - 8.75 = 8.75 meters and to calculate s2 I will do the same except this time the acceleration will be 2.8 m/s2 and initial speed = 0, correct...

This is what I got so far: F = 1.962 N a = -2.8 m/s2 ( because it's de-accelerating ) t ( to stop ) = 7 m/s / 2.8 m/s2 = 2.5s s = 7m * 2.5s * 2 ( because it will go forward and back) = 35 meters v(final) = 0 m/s does this look correct to you?

Alright, I'll see what I can come up with. Thank you very much for sacrificing your time to help me out! You've helped me out a ton. Have a nice day!

Wow, well, that was simpler than I thought! However, I'm still unsure how to get the total distance the cart traveled in those 5 seconds! The s formula you said is only used for displacement, what would be the formula for distance then?

Hmm, perhaps if velocity equals 0, cart changes the direction ( because of the load over the pulley ) ? But how would that translate into a formula?

Okay, so if the cart returned back to its original place after 5 seconds, that means he moved forward and then he returned back. But I don't know what's the distance of these two directions that the cart completed. How do I figure that out? Also, talking about the speed. Since displacement is 0...

Oh, well, I suppose it kind of makes sense! The mass over the pulley constantly pulls the wire down due to gravity, and eventually the cart runs out of energy and returns back to it's original place because of the wire. So looks like I need to use a different formula to calculate the distance...

So does that mean the acceleration will be a = -force / mass, and all the calculations will remain the same except we'll get a negative value? But if that's the case, the distance will equal to 0, because s = 7 * 5 + 0.5 * (-2.8) * 52 = 35 - 35 = 0

I'm somewhat stumped right now. So the formula for a is still FORCE / MASS , since you said all formulas remain the same. Is the force supposed to be negative now, since it opposes the thing now? But does that even make sense?

a = mass / force ? What about the distance formula? Will it remain the same? s = v(initial) * t + 0.5 * acceleration * time2

The problem stated that the cart is going to the left, but I'm unsure if that's useful. However, it said that the weight was attached to the end of the cart, so I presume the weight is supposed to slow the thing down.