Five minutes in to the trip (at t = 5 min), are the velocities of the students e
ID: 1425446 • Letter: F
Question
Five minutes in to the trip (at t = 5 min), are the velocities of the students equal? If not, which is larger? (b) For the first 5 minutes of the trip (from t = 0 to t = 5 min), are the average velocities of the students equal? If not, which is larger? (c) For the time between t = 3 min and t = 5 min is the average acceleration of student A positive, negative, or zero? (d) For the time between t = 3 min and t = 5 min is the average acceleration of student B positive, negative, or zero? 6. Elevator Prof, Lippert enters an elevator on the 3rd floor of the Metcalfe building to go up to his office on the 11 th floor. The elevator stans 8 m above the ground and begins accelerating upward at 5.0 m/s^2 for 1.0 $. Then, a disgruntled student cuts the elevator cables, causing the elevator to fall freely back toward the ground.Explanation / Answer
question 6) distance of the elevator from the ground = 8m
now, the initial velocity of the elevator, u =0 and it accelerates upwards with an acceleration of 5m/s2 for 1sec
so,
1) the velocity, v of the escalator when the cables are cut is given by :
v = u + at = 0 +5*1 = 5m/s
2) to calculate the distance of the elevator above the ground to which it travels before the rope is cut, can be determined from the equation
v2 - u2 =2as
here, s is the distance from the point where th professor enters, which is 8m above the ground and so, the total distance will be equal to (s+8)m
25-0 = 2*5*s
therefore, s = 2.5m
and the total distance above the ground, S =2.5 + 8 =10.5m
c) to calculate the velocity of the elevator when it hits the ground, we again use the relation
v2 - u2 =2aS
now, here u =5m/s, the velocity at the highest point before the rope is cut ans S = 10.5m
v2 = 25 +2*9.8*10.5
so, v =15.19m/s
d) to calculate the time taken by the elevator to hit the ground, we use the relation
v = u +at
15.19 - 5 =9.8*t
so, t = 1.04secs
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