Posted by **Help again?? Becca** on Sunday, September 17, 2006 at 10:53pm.

A small first-aid kit is dropped by a rock climber who is descending steadily at 1.9 m/s. After 2.5 s, what is the velocity of the first-aid kit, and how far is the kit below the climber?

My question is, how can you determine how fast/far the kit is going????

If the kit is dropped and under the influence of gravity, then the distance it falls is

s=(1/2)gt^2 so the distance it falls is s=(1/2)9.8m/s^2*(2.5sec)^2

The velocity of the kit is v=g*t=9.8m/s^2*2.5sec

The distance the climber has descended is 1.9m/s*2.5sec

Subtract the distance the climber descended from the distance the kit fell to see how far below the climber the kit is.

Oh.. I see. I was pretty sure there was gravity in that equation somewhere but I wasn't sure of it.

but the distance would be 31 m, and the velocity would be 25 m/s and that doesn't work. Did I mess up something with the order of operations?

Those are the number I got using those formulas too. What is it that doesn't work here?

The velocity and distance are independent of each other. Yes, the kit fell for 2.5 seconds and the climber descended for that time too. Check my second response to your first post below this one.

I see the problem. The climber was descending when the kit fell.

The distance it fell is

s=(1/2)g*t^2+v*t where g=9.8m/s^2 and v=1.9m/s

It's velocity is v=g*t+v_0 where v_0=1.9m/s

Try the calculations with those numbers.

Distance is 9.65 and velocity is 26.4? Eh?

Your ditance is wrong, but the velocity looks correct. We have

"A small first-aid kit is dropped by a rock climber who is descending steadily at 1.9 m/s. After 2.5 s, what is the velocity of the first-aid kit, and how far is the kit below the climber?"

Using s=(1/2)g*t^2+vt we have s=(1/2)9.8*(2.5)^2 + 1.9*2.5 = approx 35.4m verify this

V=gt+v_0 = 9.8*2.5+1.9 = 26.4m/s

would you subtract 1.9*2.5 since the rock climber is also descending?

No, the climber is descending, therefore the direction is down which is the direction of gravity.

oooooooh. Heh.

But I don't understand how distance and velocity are unrelated.

The formula my class is working with right now is distance= velocity*time

Yes, they are related. I may've misunderstood your comment, I'll reread it again.

Your formula distance=velocity*(time) is almost right.

Distance= averagevelocity* time is a much better formula to understand.

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