A 46.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.85 m/s in 1.07 s. It travels with this constant speed for the next 6.80 s. The elevator then slows down (undergoes a uniform acceleration in the negative y direction) for 1.95 s and comes to rest. What is the acceleration of the man while the elevator is slowing down on its way upwards? Answer in units of m/s2 and use the upward direction as positive
To find the acceleration of the man while the elevator is slowing down, we can use the equation for acceleration:
acceleration = (final velocity - initial velocity) / time
First, let's determine the initial velocity of the elevator when it starts slowing down. We know that it has been moving upwards with a constant speed of 1.85 m/s for 6.80 seconds. So, the initial velocity of the elevator is 1.85 m/s.
The final velocity of the elevator is 0 m/s since it comes to rest.
The time it takes for the elevator to slow down is given as 1.95 seconds.
Now we can calculate the acceleration:
acceleration = (0 - 1.85) / 1.95
Simplifying this, we have:
acceleration = -1.85 / 1.95
Evaluating this expression, we find:
acceleration ≈ -0.948 m/s² (rounded to three decimal places)
Since we're using the upward direction as positive, the negative sign indicates that the acceleration of the man is in the opposite direction to the motion of the elevator (downward). Therefore, the acceleration of the man while the elevator is slowing down is approximately -0.948 m/s².