A small first aid kit is dropped by a rock climber who is descending steadily at 1.3 m/s. After 2.5 s, what is the velocity of the first-aid kit, and how far is the kit below the climber?
This problem seems like it should be easy, but I'm not sure which equation to use. I'd love some help!
vi= -1.3m/s
Vf=vi-g*time
g=9.8m/s^2
d= vi*t-1/2 g t^2
I keep getting a negative answer for this...is that ok?
Yes, d is below the starting point.
It should be -26 m/s ; 31 m
To solve this problem, we can use the equations of motion. Specifically, we will use the equation for velocity and distance in terms of time.
Let's start by finding the velocity of the first aid kit after 2.5 seconds. The climber is descending steadily, so we know the velocity has a constant value. In this case, the velocity of the climber is given as 1.3 m/s.
Since the velocity is constant, we can use the formula:
velocity = initial velocity
In this case, the initial velocity is 1.3 m/s, so the velocity of the first aid kit after 2.5 seconds is also 1.3 m/s.
Next, let's find the distance the kit is below the climber. Since the velocity is constant, we can use the formula:
distance = initial position + velocity * time
The initial position is the position of the climber, which is 0 meters (since we assume the climber starts at ground level). The velocity is 1.3 m/s, and the time is 2.5 seconds.
Plugging in the values, we get:
distance = 0 + 1.3 * 2.5
distance = 3.25 meters
So, after 2.5 seconds, the first aid kit is 3.25 meters below the climber.