Recent content by The_big_dill

I do not know how you are going to solve it without any angles.

I already have velocity 1 and 2... i need distance, time or acceleration.

Interesting... I will give that a go. I already know V2.

... I am missing 2 variables, d and a. 2.52 = 02 + 2a\Deltad

V22 = V12 + 2ad

I do know 2 that are very similar. \Deltad = V1\Deltat + 1/2a\Deltat2 \Deltad = V2\Deltat - 1/2a\Deltat2 Hmmm, are you suggesting to equate them? I should end up with a quadratic.

I cannot solve for d. I do not have 'a', or delta 't'. Maybe i am missing something, but i do not see it.

I think there is some confusion here... I use "d" to represent distance V1 to represent initial velocity V2 to represent final velocity The only other equation that makes sense to me (but still missing multiple variables): V22 = V12 + 2a\Deltad I don't have distance, i also...

Exactly, thank you for clarifying that XD Also, tiny tim, i am either not familiar with that equation, or i already listed it in my original post. If you mean a = \DeltaV/\Deltat, then i don't have \Deltat, i only have t1... which is 0... which is undefined.

Homework Statement A Book (initially at rest) slides down an inclined plane, when it is at the bottom of the slope, its speed is 2.5 m/s. What is its speed half way and a third way down the slope. V1 = 0 t1 = 0 V2 = 2.5 m/s Homework Equations a = \Deltav / \Delta t v = \Deltad...

i don't know if i am missing something... but i could simplify it by just finding what \Deltad*Cos\theta is. i am simply finding the X-component, i don't think i need both components, all i need is the distance traveled in a straight line.

Let me clarify, in my original post i was attempting to make everything more brief so it is less discouraging to read. The regular stone hit the junior one at an angle, making one go 40 degrees and the other go -30 degrees, if that makes sense. I calculated a deceleration that would apply...

Sorry, i guess i had a brain fart, i am use to seeing it as a=v/t

apologize for the confusion, soothsayer is correct.

It is the angle from the horizontal (that being a straight line along the ice)