Answer to Question #135745 in Physics for Sam Far

Question

Here is the given equation to utilize for all of the questions: Vf^2=Vi^2+2aΔy.

Adam and Jamie determine that the terminal velocity for an average-sized man is 196 km/hr (54.44 m/s). This is the velocity at which the skydiver hits the seesaw. If the distance over which they slow down to a stop (velocity = 0 m/s) is 1 meter, what is the skydiver’s average acceleration?

My physics teacher had delineated to me the answer of 1400 m/s^2. As I am not quite assured how to utilize the given equation (above the question) in the context of the problem, I would greatly appreciate it if you can please demonstrate how to solve this problem with the equation, showing each step accordingly to acquire the answer of 1400 m/s^2.

Answer 1

Simply substitute the values into the equation:

Final velocity: Vf=0 m/s

Initial velocity: Vi=54.44 m/s

Distance they had to change speed from 54.44 m/s to 0 m/s: Δy=1 m.

Now use the equation (do not forged about squares!)

$$0=54.44^2+2a\cdot1,\\ a=-\frac{54.44^2}{2\cdot1}=-1482\text{ m/s}^2.$$

Negative sign means that they decelerated along this 1 meter.