A small first aid kit is dropped by a rock climber who is descending at a constant velocity of
- 1.3m/s.
a) Sketch the velocity vs. time graph for both the climber and her first aid kit and paste your sketch in the space below.
b) After 1.5 seconds, what is the velocity of the first-aid kit?
c) How far is the kit below the climber?
a) To sketch the velocity vs. time graph, we need to consider that the rock climber is descending at a constant velocity of -1.3 m/s. This means that the velocity of the climber will remain constant throughout the time. The first aid kit, on the other hand, will also descend but without any external force acting on it (such as the climber pulling it). This means that the kit will experience a free fall and its velocity will increase over time due to gravity.
For the climber, the velocity vs. time graph will be a horizontal line at -1.3 m/s since the velocity is constant.
For the first aid kit, the velocity vs. time graph will be a straight line with a positive slope because the velocity is increasing due to gravity's pull. The slope represents the rate of change of velocity over time. The graph will start from the initial velocity of 0 m/s (when it is dropped) and continuously increase.
b) After 1.5 seconds, the velocity of the first aid kit can be determined using the formula for free fall velocity. The formula is:
v = gt
where v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Plugging in the values:
v = 9.8 m/s^2 * 1.5 s = 14.7 m/s
So, after 1.5 seconds, the velocity of the first aid kit is 14.7 m/s.
c) To calculate how far the kit is below the climber, we need to use the distance formula for free fall:
d = (1/2) * g * t^2
where d is the distance, g is the acceleration due to gravity, and t is the time.
Using the values:
d = (1/2) * 9.8 m/s^2 * (1.5 s)^2 = 11.025 meters
So, the first aid kit is approximately 11.025 meters below the climber.