Recent content by Fisherman166

Try using the same equation you used to solve part A, only solve for the initial velocity this time. V2= 0 m/s deltaX = 35m (or maybe 34m so you wouldn't technically touch the deer?) a = -10 m/s^2 That doesn't really take into account the 0.5sec to brake, but I don't know how to account for that.

Alright, I resolved and got T1=3.01973 sec and T2=-0.130 sec Then I did: deltaX = 4.9(3.01973)^2 = 44.7m Is 44.7m actually how deep the well is?

I solved for T1 = 3.15 - T2 and T2 = 3.15 - T1. I then substituted them into the first formula and solved for each. Which gave answers of: T1=3.31 sec and T2=-0.156 sec Which confuses me a bit. How can I have a negative value for T2?

I'm still having a hard time visualizing how those two equations are helpful. Am I suppose to combine the two together somehow (maybe by replacing t1 and t2 with just T?)?

That is where I am getting tripped up at though. Since I only know the initial velocity and the acceleration for the rock, all of my formulas have 2 unknowns in them, which means I can't solve for time.

So what you're saying is that the equation should actually look like: 1/2(-9.8)(t1^2) = 343(t2) since sound won't be affected by gravity?

My pleasure! I'm glad I could help out. This is also my first week of physics with calculus, so this was a good exercise to make sure I could use my constant acceleration equations correctly :D

Hmm...Nothing really comes to mind from it other than it looks similar to 1/2(-9.8)(t1^2) on the other side of the equation. Sorry if I'm being a bit slow on this.

I'm not sure I completely follow. Do you mean do something like: 0(t1) + 1/2(-9.8)(t1^2) = 343(t2) + 1/2(-9.8)(t2^2) Where each of the equations should equal the same displacement? So then I should solve for t1 and t2?

That's what I got as my answer to the problem.

The next question is asking for the displacement (or how far the sled traveled in meters) of the sled after 1.6 seconds. The sled couldn't have stayed in place if it's velocity increased to 397 m/s after 1.6seconds, so it has to have a displacement. Trying using the equation: deltaX = Vo(t) +...

397 m/s would be your final velocity.

The acceleration is constant, but it is not zero. If it was 0, it would not be possible for the sled to get to a velocity of 397 m/s in 1.6s from rest (0 m/s). Trying using the equation V = Vo + a(t) and solve for a.

Yes, because speed is the magnitude of the velocity. If the magnitude of the velocity never changes, then the acceleration is 0 m/s^2.

v2-v1/t2-t1 is just the slope formula y2-y1/x2-x1 Since acceleration is the slope (or the derivative) of velocity, finding the slope between two points of velocity will give you average acceleration. Your equation is pretty much the exact same as that, only your t doesn't bother with...