Question

So I need to find the instantaneous velocity of a falling rock at 4 seconds in its fall. At 0 seconds the rock is 80 m high, and at 10 seconds the rock is 0 m high. I feel like I'd need the function for the position of rock at a specific time to find the instantaneous velocity, but I can't find the function, and I can't think of how to find the velocity any other way. I'd really appreciate an explanation. I'm also new here, and I'm not sure how to 'give a medal' but I'll give a medal as soon as I know how to. If you guys tell me how to give a medal then it's easier. Thank you ahead of time!

Answer 1

maybe 80/10 is 8 so at 1 second it goes down 8 seconds so 8*4 is 32 and 80-32 is 44

Answer 2

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Answer 3

I think you might be onto something, but I don't understand exactly what you're telling me. Btw, I forgot to add this in, but I know for a fact that the answer is -7.0 m/s^2, I just don't know how to get there. Thank you again :P :]

Answer 4

I'm hoping this is how tagging works @NetherCreep333

Answer 5

im in 7th grade so i barely even know calculus im suprised im eb=ven on to something

Answer 6

Well you're onto the same thing I am xD I'm going to keep trying and I'll tell you if I figure it out. I'll type it in if I do.

Answer 7

It is a tough question for seventh grade, and it is primarily a physics, not a calculus question. As you state the question, that the rock is at 0 meters after 10 seconds makes me think that someone is trying to fool you, because it also could have been at 0 meters at nine seconds. The velocity is related to its acceleration times the time it is accelerating. The acceleration due to gravity is 9.8 m/s^2. The time is 4 seconds. velocity is equal to acceleration times the time (undergoing acceleration). At 4 seconds, it is 9.8 times 4, with the units coming out as meters/second. After 10 seconds, your rock would have fallen much farther than 80 meters, by the way, had the ground not been in the

Answer 8

"that the rock is at 0 meters after 10 seconds makes me think that someone is trying to fool you." That is not necessary. The rock might as well have been thrown from beneath a hole. Either ways, the info about the rock being 80 m at the beginning gives us all we need. if acceleration was 10 m/s^2, you'd have instantaneous velocity being 10t + v_initial. distance would then be 0.5*10*t + v_initial*t + d_verticalShift.
the vertical shift was 80 meters up, which is just how high the ball is or where we "start." the fall. We don't know what the initial velocity is, yet. but from the displacement function, (you were given 2 values!) you can find it easily.
displacement(t) = 5t + v_initial*t + 80
At t = 10.... we are at displacement 0 m.
0 = 50 + v_initial * 10 + 80
v_initial = -13 m/s
v_instant(t) = 10t -13
this is all based on the assumption that acceleration is 10 m/s^2, though. Could be 9.8 m/s^2, could be something else, so you might need to approach this a little bit differently.
Sorry. this is so long...

Answer 9

The distance the rock falls under the force of gravity, given an initial velocity of 0, is 1/2 at^2 - that is 1/2(9.8)(100), given the ten second mark. That is 490 meters. It only has 80 meters to fall. At 4 seconds it will be 78.4 meters. Since the ground is at 80 meters, it will still be moving, so the velocity will not be zero. Immediately after though, it will hit the ground and the equations describing its motion will change.

The 80 meter distance limit explains why it is found at 80 meters at 10 seconds, under the force of gravity. It will be at 80 meters at 20, 30 and ever after because it is not moving any longer. The question is trying to mis-direct attention and get a student to answer by using a simple linear equation to estimate velocity, rather than using the quadratic equation that describes distance and velocity under the influence of gravitation. I won't solve the velocity side of the equation because that is his problem. I only wanted to point out the trick in the question.

The rock is falling, as the problem states, not rising. There is no reason to believe that the rock is falling through a hole, nor that it has an initial velocity. If it did have an initial velocity there would be no way to determine what the velocity was at any given time without revealing that initial velocity.

Answer 10

Sorry guys, turns out this problem was one of those where you just guess by looking at a graph. You actually CAN'T solve this with math. If you somehow managed to then wow, and congratulations.

Answer 11

Sorry guys, turns out this problem was one of those where you just guess by looking at a graph. You actually CAN'T solve this with math. If you somehow managed to then wow, and congratulations.

Answer 12

You have been misinformed, if you have adequately stated the problem. The graph may have been constructed by observation of the fall of the rock, but under uniform acceleration, it "also" could have been entirely plotted according to a mathematical equation describing velocity as a function of time.