A small first-aid kit is dropped by a rock

climber who is descending steadily at 1.3 m/s.
After 4.0 s, what is the velocity of the
first-aid kit? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m/s
012 (part 2 of 2) 10.0 points
How far is the kit below the climber after the
4.0 s?
Answer in units of m

To find the velocity of the first-aid kit after 4.0 seconds, you can use the formula:

velocity (v) = initial velocity (u) + acceleration (a) * time (t)

The initial velocity of the first-aid kit is 0 m/s since it was dropped. The acceleration acting on the kit is due to gravity and is -9.81 m/s^2 (negative because it is acting in the opposite direction of the descent). The time is given as 4.0 seconds.

Plugging these values into the formula:

v = u + a * t
= 0 + (-9.81) * 4.0
= -39.24 m/s

Therefore, the velocity of the first-aid kit after 4.0 seconds is -39.24 m/s.
(Note: The negative sign indicates that the kit is moving downward.)

To find how far the kit is below the climber after 4.0 seconds, you can use the formula:

distance (s) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * time^2

Here, the initial velocity is again 0 m/s, the time is 4.0 seconds, and the acceleration is -9.81 m/s^2.

Plugging these values into the formula:

s = u * t + 0.5 * a * t^2
= 0 * 4.0 + 0.5 * (-9.81) * (4.0)^2
= 0 + 0.5 * (-9.81) * 16.0
= 0 + (-78.48)
= -78.48 m

Therefore, the kit is 78.48 meters below the climber after 4.0 seconds.
(Note: The negative sign indicates that the kit is below the climber.)