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.
the velocity is -24.65. the distance is 30.66 m
but answeer in the book says that the velocity is -27m/s and the distance is 31m
I apologize for any confusion. Distance and velocity are indeed related. The formula you mentioned, distance = velocity * time, is correct. In this scenario, the initial velocity of the first-aid kit is 0 because it was dropped. As it falls under the influence of gravity, its velocity increases over time. The formula for calculating the distance it falls in this case is:
distance = (1/2) * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time in seconds.
To calculate the velocity of the first-aid kit after 2.5 seconds, we can use the formula:
velocity = g * t
where t is again the time in seconds.
For the distance below the climber, we subtract the distance the climber descended from the distance the kit fell:
distance below climber = distance - (velocity of climber * time)
Substituting the given values into these formulas will give you the specific answers for this scenario.
Let me clarify the relationship between distance and velocity. In general, distance and velocity are related, but they are not always directly proportional. Distance is the total length traveled, while velocity is the rate at which an object moves in a certain direction. Velocity takes into account both the speed of the object and its direction.
In the formula distance = velocity * time, you are correct that this formula relates distance and velocity, but it assumes a constant velocity. It is important to consider that velocity can change over time, which means the relationship between distance and velocity may not always be straightforward.
In the case of the rock climber dropping the first-aid kit, the velocity is not constant because the climber is descending steadily at a speed of 1.9 m/s. This means that as the kit is dropped, it is also affected by gravity, which adds an additional downward velocity. As a result, the velocity of the kit will be the sum of the initial velocity of the climber (1.9 m/s) and the velocity due to gravity (9.8 m/s^2 * 2.5 s).
To calculate the distance the kit falls, we use the formula s = (1/2)gt^2 + vt, where s represents the distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s^2), t is the time (2.5 s), and v is the initial velocity of the kit (1.9 m/s).
Using this formula, we find s = (1/2)(9.8 m/s^2)(2.5 s)^2 + (1.9 m/s)(2.5 s) = approximately 35.4 m.
To calculate the velocity of the kit, we use the formula v = gt + v0, where v represents the final velocity, g is the acceleration due to gravity, t is the time, and v0 is the initial velocity of the kit (1.9 m/s).
Using this formula, we find v = (9.8 m/s^2)(2.5 s) + 1.9 m/s = approximately 26.4 m/s.
Therefore, after 2.5 seconds, the first-aid kit is approximately 35.4 meters below the climber and has a velocity of approximately 26.4 m/s.