A small first-aid kit is dropped by a rock climber who is descending steadily at 1.6 m/s.
(a) After 3.5 s, what is the velocity of the first-aid kit?
(b) After the 3.5 s, how far is the kit below the climber?
(a) v = 1.6 m/s + (3.5 s * g)
(b) d = 3.5 s [1.6 m/s + (1/2 g * 3.5 s)]
what is g?
g is the gravitational acceleration
9.81 m/s^2
so a is 3.5 times 9.81 then add 1.6?
To answer these questions, we need to understand the basic principles of motion, specifically the concept of velocity and distance. Velocity is a measure of how fast an object is moving in a particular direction, whereas distance refers to the total amount of ground covered by an object.
(a) To determine the velocity of the first-aid kit after 3.5 seconds, we need to know its initial velocity and the acceleration acting upon it. In this case, we are told that the rock climber is descending steadily at a velocity of 1.6 m/s. Since the first-aid kit is dropped by the climber, it will have the same initial velocity.
Velocity is a vector quantity, which means it has both magnitude and direction. In this case, the direction of the velocity is negative since the kit is descending. So, the initial velocity of the first-aid kit is -1.6 m/s.
After 3.5 seconds, the velocity of the first-aid kit can be found by multiplying the acceleration (which remains constant at 0 m/s² since the kit is in free fall) by the time and adding it to the initial velocity:
Velocity = Initial Velocity + (Acceleration * Time)
In this case, the equation becomes:
Velocity = -1.6 m/s + (0 m/s² * 3.5 s)
Since the acceleration is zero, the velocity after 3.5 seconds will be equal to the initial velocity of -1.6 m/s. Therefore, the velocity of the first-aid kit remains at -1.6 m/s after 3.5 seconds.
(b) To find how far the kit is below the climber after 3.5 seconds, we can use the equation for distance traveled during uniform motion:
Distance = Initial Velocity * Time + (0.5 * Acceleration * Time²)
In this case, since the kit is in free fall, the acceleration will be the acceleration due to gravity, approximately 9.8 m/s². The initial velocity of the kit is -1.6 m/s, and the time is 3.5 seconds. Plugging these values into the equation, we get:
Distance = -1.6 m/s * 3.5 s + (0.5 * 9.8 m/s² * (3.5 s)²)
Simplifying the equation, we have:
Distance = -5.6 m + (0.5 * 9.8 m/s² * 12.25 s²)
Distance = -5.6 m + 59.95 m
Distance ≈ 54.35 m
Therefore, after 3.5 seconds, the first-aid kit is approximately 54.35 meters below the climber.